Cathal Bowe - PhD

DYNAMIC INTERACTION OF TRAINS AND RAILWAY

BRIDGES USING WHEEL RAIL CONTACT METHOD

by

CATHAL BOWE

National University of Ireland, Galway

Faculty of Engineering

Department of Civil Engineering

A thesis submitted in partial fulfilment of the requirements

for the degree of Doctor of Philosophy

Dean of College of Engineering & Informatics Research Supervisor

Prof. Padraic E. Donoghue Dr. Thomas Mullarkey

November 2009

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ABSTRACT

The primary objective of this thesis is to develop numerical models that can be used in the safety

assessments of rail-bridge systems. Iarnród Éireann, co-sponsor of this research, is particularly

interested in the dynamic effects caused by the increasing axle loads and by the increase in line speeds

of trains travelling over bridges. The centre span of Boyne Viaduct Railway Bridge is investigated as a

case study, including actions of the railway tracks leading up to and away from the bridge, and is

modelled as a two-dimensional and three-dimensional truss.

ANSYS is the finite element program used throughout the thesis to analyse the dynamic behaviour of a

train traversing a railway bridge. However, it was discovered that this program has many limitations; in

particular, its contact elements are unable to correctly model track irregularities or a train braking.

Nevertheless, this problem is overcome by the author’s development of his own wheel-rail contact

element, which can model the vertical, longitudinal and lateral responses of each wheel on the rail and

also includes track irregularities and wheel-rail separation. This development assumes that a Hertzian

spring element exists between each wheel and the rail and is often called a sprung mass wheel system.

Under smooth rail conditions, results show that the author’s wheel-rail contact element performs better

than the commercial node-to-surface contact elements in ANSYS because the author’s system

maintains accuracy when the number of elements in the model is reduced unlike the ANSYS contact

element system which loses its accuracy.

In this thesis, the author also develops both a modal and finite element model of a moving unsprung

mass traversing a bridge that can include track irregularities. The author highlights the similarities and

differences between the modal and finite element model from the point of view of the form of the final

matrices. The unsprung wheel system assumes that the wheel is permanently attached to the rail and

cannot separate from it; thus, the unsprung mass experiences both local and convective velocities and

accelerations, which must be taken into account. Early studies of the moving unsprung mass show that

the convective velocity and acceleration were omitted from the model; thus their solution is inaccurate.

Nevertheless, many authors are still comparing the results of their models with this inaccurate solution,

ignoring the issue of convective acceleration. The author addresses this issue in the thesis by

specifically presenting the correct solution of a moving unsprung mass traversing a cantilever beam.

Other results reveal that the developed unsprung systems are comparable with the developed wheel-rail

contact element when the Hertzian stiffness is reasonably large value and separation is not allowed.

From our case study of the centre span of the Boyne Viaduct Railway Bridge, the author found that a

passenger train travelling at approximately 200 km/hr has a dynamic deflection equal to the static

loaded deflection multiplied by a factor of 1.07. Moreover, it was found that the bridge could be

susceptible to resonance generated by repetitive loaded vehicles travelling at service speeds; namely,

the train Type 1 travelling at 228 km/hr or train Type 8 travelling at 110 km/hr from the Eurocodes.

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DECLARATION

I hereby declare that this thesis, in whole or in part, has not been previously submitted

to any other university as an exercise for a degree. Except where specific reference to

the work of others is given in the text, this thesis is entirely my own work.

Cathal Bowe, November 2009

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ACKNOWLEDGEMENTS

This thesis describes work conducted at the Department of Civil Engineering of the National

University of Ireland, Galway (NUIG). The research project has been generously funded by

Irish Rail and Enterprise Ireland. This financial support is greatly acknowledged.

The research was carried out under the supervision of Dr. Thomas Mullarkey, to whom I wish

to express my sincere gratitude for his assistance throughout this thesis. His knowledge of

structural dynamics, elastic beam theories, and finite element methods as well as his help

developing the many mathematical formulations within this thesis are greatly appreciated, in

particular, his help with the development of the time varying moving forces in Chapter 3, the

wheel-rail contact element in Chapter 4, the modal and finite element method of a moving

unsprung mass in Chapter 5, the irregularity functions in Chapter 6 and the many

mathematical equations found in my Appendices. I would also like to thank him for the

countless hours that he has spent reading and fine tuning my thesis.

I would like to thank Dr. Michal Majka, a good friend, for his help, encouragement, and

dynamic expertise. From my early PhD days to its final stages, Michal has always been

supportive, for which, I’m very grateful. In addition, I would like to thank Dr. Michael

Hartnett for his dynamic expertise and for securing financial support in the third year of my

studies.

I would like to thank my parents, Charlie and Teresa Bowe, for their financial support and

their constant encouragement over the years. Additionally, I would like to thank brothers,

sisters and friends, in particular my finacée Amanda, for their support.

Finally, I wish to thank all of my colleagues and staff at NUIG and those who have not been

mentioned by name, but who have helped me during the course of the work,

Thank you.

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CONTENTS

LIST OF SYMBOLS AND ABBREVIATIONS .................................................... xii

INTRODUCTION ....................................................................................................... 1

1.1 Background & Motivation ............................................................................... 1

1.2 ANSYS Strength & Weaknesses ..................................................................... 3

1.3 Chapter Summary ............................................................................................ 6

LITERATURE REVIEW ......................................................................................... 11

2.1 Introduction .. ................................................................................................ 11

2.2 Modelling Railway Vehicles ......................................................................... 11

2.2.1 History of Railway Vehicle Dynamics ................................................ 11

2.2.2 Vehicle Model ...................................................................................... 13

2.2.2.1 Moving Constant Force .......................................................... 13

2.2.2.2 Weight and Mass of Moving Wheel ...................................... 14

2.2.2.3 Mass of the Moving Vehicle .................................................. 15

2.2.3 Vehicle Characteristics ......................................................................... 16

2.2.3.1 Vehicle Spacing ..................................................................... 16

2.2.3.2 Vehicle Speed ........................................................................ 17

2.2.3.3 Braking & Accelerating of Vehicles ...................................... 18

2.2.3.4 Passenger Riding Comfort ..................................................... 19

2.2.3.5 Ratio between Sprung & Unsprung Masses .......................... 20

2.2.3.6 Ratio between the Vehicle Mass & Bridge Mass .................. 20

2.3 Modelling Railway Bridges ........................................................................... 21

2.3.1 Railway Bridge Types .......................................................................... 21

2.3.1.1 Beam Railway Bridge ............................................................ 21

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2.3.1.2 Truss Railway Bridge ............................................................. 22

2.3.1.3 Suspension & Cable Bridges .................................................. 22

2.3.1.4 Boyne Viaduct Railway Bridge .............................................. 23

2.3.2 Bridge Characteristics .......................................................................... 24

2.3.2.1 Bridge Natural Frequency ...................................................... 24

2.3.2.2 Bridge Damping ..................................................................... 25

2.3.2.3 Bridge Span Length ................................................................ 26

2.3.2.4 Rail Approaches ..................................................................... 27

2.3.2.5 Dynamic Amplification Factor ............................................... 27

2.3.3 Numerical Bridge Solutions ................................................................. 28

2.3.3.1 Finite Element Model ............................................................. 29

2.3.3.2 Method of Modal Superposition ............................................ 30

2.4 Modelling Irregularities ................................................................................. 31

2.4.1 Wheel Irregularities .............................................................................. 31

2.4.2 Rail Irregularities .................................................................................. 31

2.5 Wheel-Rail Contacts ...................................................................................... 32

2.5.1 ANSYS contact elements ..................................................................... 33

2.5.2 Hertz Contact Theory ........................................................................... 34

2.5.3 Vehicle-bridge interaction element ...................................................... 34

2.6 Summary ....... ................................................................................................ 35

WHEEL FORCE REPRESENTED AS TIME VARYING NODAL FORCES

AND MOMENTS ...................................................................................................... 37

3.1 Introduction ... ................................................................................................ 37

3.2 Single Moving Force as a Function of Time ................................................. 40

3.2.1 Development of single moving force ................................................... 40

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3.2.1.1 Moving force using an approximate simple model ................ 44

3.2.1.2 Moving force using an exact numerical model ...................... 48

3.2.1.3 Moving single force using ANSYS contact elements ............ 52

3.2.2 Validation of moving force as a function of time ................................. 53

3.2.2.1 Cantilever beam subjected to a moving load ......................... 53

3.2.2.2 Simply supported beam subjected to a moving load ............. 54

3.2.2.3 Sensitivity analysis of the developed systems ....................... 56

3.2.2.4 Examining internal forces of the developed systems ............. 60

3.2.3 Application of single load traversing Boyne Viaduct .......................... 73

3.3 Multiple Moving Forces as a Function of Time ............................................... 78

3.3.1 Development of multiple moving forces .............................................. 78

3.3.1.1 Simple solution without overlapping time functions ............. 78

3.3.1.2 Simple solution with overlapping time functions .................. 84

3.3.1.3 Multiple moving forces using ANSYS contact elements ...... 87

3.3.2 Validation of multiple moving forces as a function time ..................... 88

3.3.3 Application of multiple forces traversing the Boyne Viaduct .............. 96

3.3.3.1 Railway Vehicles ................................................................... 96

3.3.3.2 Two-Dimensional Boyne Bridge ........................................... 97

3.3.5.3 Three-Dimensional Boyne Bridge ....................................... 101

3.3.5.4 Twin-track Railway Bridge .................................................. 106

3.3.5.5 Boyne Viaduct subjected to Eurocodes (1991) rail loads .... 109

3.4 Discussion of results and Conclusion ............................................................. 113

SPRUNG MASS REPRESENTED BY TIME VARYING STIFFNESS

MATRICES .............................................................................................................. 117

4.1 Introduction .. .............................................................................................. 117

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4.2 Development of the Wheel-Rail Contact Elements ..................................... 120

4.2.1 Vertical Spring Element using time varying matrices ........................ 120

4.2.2 Longitudinal Spring Element using time varying matrices ................ 128

4.2.3 Lateral Spring Element using time varying matrices ......................... 135

4.2.4 Multiple wheels using the WRC element ........................................... 144

4.3 Validating the Wheel-Rail Contact Elements .............................................. 146

4.3.1 Wheel as a sprung load traversing a cantilever beam ......................... 146

4.3.2 Wheel as a sprung mass traversing a cantilever beam ........................ 148

4.3.3 Wheel as a sprung load traversing a simply supported beam ............. 151

4.3.4 Wheel as a sprung mass traversing a simply supported beam............ 153

4.3.5 A travelling bouncing wheel traversing a rigid rail and beam ........... 156

4.3.6 Sprung & unsprung systems at a wide range of contact stiffness ...... 158

4.3.7 Sensitivity analysis of the WRC element ........................................... 160

4.3.8 Simply supported beam subjected to a two-wheeled system ............. 162

4.3.9 Two-wheeled vehicle subjected to braking effects ............................ 165

4.4 Application of WRC element with Boyne Viaduct ..................................... 170

4.4.1 Single sprung wheel traversing two-dimensional Boyne Viaduct ..... 170

4.4.2 Railway Vehicle ................................................................................. 175

4.4.3 Two-dimensional bridge-train model ................................................. 179

4.4.4 Braking and accelerating effects of train on the bridge ...................... 183

4.4.5 Three-dimensional bridge-train model ............................................... 188

4.4.6 Effects of lateral cross winds on the train as it travels ....................... 192

4.5 Discussion of results and Conclusions ......................................................... 194

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UNSPRUNG MASS REPRESENTED BY TIME VARYING MASS, DAMPING,

AND STIFFNESS MATRICES WITHIN A MODAL AND FINITE ELEMENT

FRAMEWORK ........................................................................................................ 197

5.1 Introduction .............................................................................................. 197

5.2 Development of the unsprung wheel system ............................................... 200

5.2.1 Modal superposition model incorporating moving unsprung mass ... 200

5.2.1.1 Modal superposition model of a moving load ........................ 205

5.2.1.2 Unsprung wheel-rail separation ............................................. 205

5.2.2 Unsprung finite element solution for unsprung moving mass ........... 207

5.2.3 Modal superposition multiple unsprung masses traversing beam...... 212

5.3 Validating the Unsprung Mass Systems ...................................................... 216

5.3.1 Wheel as a moving unsprung load traversing a cantilever beam ....... 216

5.3.2 Wheel as a moving unsprung mass traversing a cantilever beam ...... 218

5.3.3 Moving unsprung system traversing a simply supported beam ......... 224

5.3.4 Unsprung wheel system at wide range of speeds ............................... 229

5.3.5 Sensitivity analysis of the unsprung systems ..................................... 230

5.3.6 Multiple unsprung vehicles traversing a simply supported beam ...... 236

5.4 Application to the Boyne Viaduct ............................................................... 238

5.4.1 Unsprung wheel traversing Boyne Viaduct modelled as a beam ....... 238

5.4.2 Unsprung wheel traversing two-dimensional Boyne Viaduct ........... 242

5.4.3 Multiple vehicles traversing two-dimensional Boyne Viaduct .......... 247

5.4.4 Multiple vehicles traversing three-dimensional Boyne Viaduct ........ 251

5.5 Discussion of results and Conclusions ........................................................ 255

WHEEL RAIL SYSTEMS ON IRREGULARITIES .......................................... 259

6.1 Introduction ................................................................................................ 259

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6.2 Development of Irregularity models ............................................................ 261

6.2.1 Sprung wheel incorporating rail irregularities – WRC model ............ 261

6.2.2 Unsprung wheel incorporating rail irregularities – Modal model ...... 264

6.2.3 Unsprung wheel incorporating rail irregularities – FEM model ........ 268

6.2.4 Sprung wheel incorporating lateral rail irregularities ......................... 271

6.3 Validating systems with irregularities ......................................................... 274

6.3.1 Static analysis of rigid rail with irregularities .................................... 275

6.3.2 Transient analysis of beam and rigid rail with irregularities .............. 275

6.3.3 Manchester Benchmark simulation using author’s models ................ 280

6.4 Boyne Viaduct with irregularities along its rails ......................................... 288

6.4.1 Train model undergoing in-phase irregularities on each rail .............. 289

6.4.2 Train model undergoing out-of-phase irregularities on each rail ....... 295

6.4.3 Train model undergoing random irregularities on each rail ............... 301

6.4.4 Boyne Viaduct with lateral irregularities in-phase on each rail ......... 306

6.5 Discussion of results and Conclusions ......................................................... 311

CONCLUSIONS & RECOMMENDATIONS ...................................................... 313

7.1 Thesis Summary & Conclusions .................................................................. 313

7.2 Recommendations for future work .............................................................. 318

CONVENTION & ELASTIC BEAM THEORY .................................................. 319

A.1 Introduction ... .............................................................................................. 319

A.2 Conventions .............................................................................................. 319

A.2.1 Convention for coordinate axes ......................................................... 319

A.2.2 Convention for moments .................................................................... 320

A.2.3 Convections for internal forces and moments of a beam ................... 321

A.2.4 Convention for stresses ...................................................................... 322

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A.3 Moment-Curvature Relationship ................................................................. 323

A.4 Differential equations governing the transverse deformation of a beam ..... 328

A.5 Differential equations governing the longitudinal deformation of a beam .... 333

A.6 Differential equations governing the tensional deformation of a beam ...... 336

FEM APPLIED TO EQUATION OF MOTION OF A BEAM .......................... 341

B.1 Introduction .. .............................................................................................. 341

B.2 Beam element in the x-y plane ..................................................................... 341

B.3 Beam element in the x-z plane ..................................................................... 350

B.4 Beam element along x-axis .......................................................................... 357

B.5 Beam element subjected to torque about the x-axis .................................... 362

B.6 Structural beam element .............................................................................. 365

B.7 Transformation from local to global axes.................................................... 367

B.8 Rotating the element equation from local to global coordinates ................. 374

B.9 Equilibrium of the joints – Assembly .......................................................... 377

B.10 Applying rotations to axial elements ........................................................... 380

B.11 Equation for the axial extension of a spring element .................................. 384

NATURAL FREQUENCIES AND MODAL SHAPES FOR A BEAM ............. 387

C.1 Mode shape for any beam ............................................................................ 387

C.2 Cantilever beam – Natural Frequencies and Mode Shapes ......................... 391

C.3 Fixed-Fixed beam – Natural Frequencies and Mode Shapes ...................... 397

C.4 Simply supported beam – Natural Frequencies and Mode Shapes ............. 401

C.5 Dimensionless speed ratio α for a simply supported beam ......................... 405

EQUATION FOR DAMPING ................................................................................ 413

D.1 Introduction .. .............................................................................................. 413

D.2 Viscous damping ......................................................................................... 413

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D.3 Numerical damping ...................................................................................... 419

NEWMARK TIME INTEGRATION SCHEME .................................................. 423

E.1 Introduction ... .............................................................................................. 423

E.2 Newmark time integration ........................................................................... 423

E.3 Defining the parameter γ and β .................................................................... 427

BOYNE VIADUCT - TECHNICAL PARAMETERS ......................................... 429

F.1 Dimensions & Geometrical Properties ........................................................ 429

F.2 Static analysis of an unloaded Boyne Viaduct ............................................. 434

F.3 Representing a truss as a simply supported beam ........................................ 434

F.4 Modal analysis of the Boyne Viaduct railway bridge .................................. 436

RAILWAY VEHICLE DYNAMICS ..................................................................... 439

G.1 Introduction ... .............................................................................................. 439

G.2 Axle spacing and weights as a moving force ............................................... 440

G.3 Axle positioning for maximum loading of Boyne Bridge ........................... 442

G.4 Exact model with overlapping time functions ............................................. 444

G.5 Vehicle dimensions and parameters............................................................. 447

G.6 Modal analysis of the railway vehicles ........................................................ 457

REFERENCES ....................................................................................................... 463

BIBLIOGRAPHY .................................................................................................... 472

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LIST OF SYMBOLS AND ABBREVIATIONS

The following is the list of symbols and abbreviations used throughout this thesis. All

the symbols are defined at the place they appear in the text. Vector-matrices are

defined by bold letters; bold lower case letters indicate local coordinate system, while

bold upper case letters indicate global coordinate system. A dot placed over a quantity

denotes the derivative with respect to time variable t, while a dash attached to a

quantity denotes the derivative with respect to spatial variable x.

Upper Case Times New Roman

A Area

wA distance between two axle of a single bogie

wB distance between bogies of a single carriage

SC global damping matrix of the entire structure

wC distance between the rear wheel of one carriage and the front

wheel of the following carriage behind it

wD distance between the front wheels of consecutive carriages

E young’s modulus of Elasticity

xE unit vector parallel to the global x-axis

yE unit vector parallel to the global y-axis

zE unit vector parallel to the global z-axis

bF force applied to a bogie

wF force applied to a wheel

xiii

vF force applied to a vehicle body

LxjF global horizontal force in Hertzian spring at local node;j = 1, 2

LyjF global vertical force in Hertzian spring at local node; j = 1, 2

( )F t force due to the unsprung mass

F global force vector of a single finite element

SF global force vector of the entire structure

G weight of the bridge

G shear modulus of Elasticity

( )jG χ rotational beam element shape functions at local node; j = 1, 2

( )jG χ′ 1st derivative of the rotational beam element shape functions with

respect to χ at local node; j = 1, 2

( )jG χ′′ 2nd derivative of the rotational beam element shape functions

with respect to χ at local node; j = 1, 2

( )jG χ′′′ 3rd derivative of the rotational beam element shape functions with

respect to χ at local node; j = 1, 2

( )jH χ linear shape functions at local node; j = 1, 2

( )xH Heaviside function

I second moment of inertia

I unit vector

pI polar moment of area about the y-axis

vI vehicle body mass moment of inertia

yI second moment of area about the y-axis

xiv

zI second moment of area about the z-axis

K global stiffness matrix of a single finite element

SK global stiffness matrix of the entire structure

11LK Upper left corner quadrant of the global (3x3) stiffness matrix

2 11LK Upper left corner quadrant of the global (2x2) stiffness matrix

2 12LK Upper right corner quadrant of the global (2x2) stiffness matrix

2 21LK Bottom left corner quadrant of the global (2x2) stiffness matrix

2 22LK Bottom right corner quadrant of the global (2x2) stiffness matrix

L total length of the beam

M bending moment

M global mass matrix of a single finite element

SM global mass matrix of the entire structure

ˆjM bending moment of element at local node; j = 1, 2

bM bogie mass

vM vehicle body mass

wM wheel mass

xM bending moment about the x-axis

yM bending moment about the y-axis

ˆyjM bending moment about y-axis at local node; j = 1, 2

zM bending moment about the z-axis

ˆzjM bending moment about z-axis at local node; j = 1, 2

cN number of railway carriages

xv

( )jN χ transverse beam element shape functions at local node; j = 1, 2

( )jN χ′ 1st derivative of the transverse beam element shape functions


( )jN χ′′ 2nd derivative of the transverse beam element shape functions


( )jN χ′′′ 3rd derivative of the transverse beam element shape functions


P moving point force

P axial force parallel to the x-axis

1P point force from the front wheel of the first carriage

jP point force from the front wheel of carriage j

ˆjP axial force in the x-direction at local node; j = 1, 2

xP global x components of a unit vector

yP global y components of a unit vector

zP global x components of a unit vector

Q shear force

jQ point force from the second wheel of carriage j

ˆjQ shear force of element at local node; j = 1, 2

yQ shear force parallel to the y-axis

ˆyjQ shear force in y-direction at local node of element; j = 1, 2

zQ shear force parallel to the z-axis

ˆzjQ shear force in z-direction at local node of element; j = 1, 2

xvi

jR point force from the third wheel of carriage j

( )nR t transverse and rotational displacement; n = 1 to 4

( )nR t& transverse and rotational velocity; n = 1 to 4

( )nR t&& transverse and rotational acceleration; n = 1 to 4

0R rotation matrix

Rα rotation α about the y-axis

Rβ rotation β about the z-axis

Rγ rotation γ about the x-axis

jS point force from the fourth wheel of carriage j

T moment about the x-axis i.e. torque

ˆjT torque about x-axis at local node of element; j = 1, 2

T transformation matrix

jU global displacement in x-direction at local node; j = 1, 2

jU&& global acceleration in x-direction at local node; j = 1, 2

LxjU global displacement of the Hertzian spring in the x-direction

at local node; j = 1, 2

LyjU global displacement of the Hertzian spring in the y-direction at

local node; j = 1, 2

U global nodal displacement vector of a single finite element

SU global displacement vector of the entire structure

SU& global velocity vector of the entire structure

U&& global nodal acceleration vector of a single finite element

xvii

SU&& global acceleration vector of the entire structure

jV global displacement in y-direction at local node; j = 1, 2

jV&& global acceleration in y-direction at local node; j = 1, 2

jW global displacement in z-direction at local node; j = 1, 2

jW&& global acceleration in z-direction at local node; j = 1, 2

X global coordinate

( )X t horizontal position of the unsprung mass at time t

( ), ,m j k

X t horizontal position of the m-th wheel of the k-th bogie

of the j-th carriage of the train at time t

Y global coordinate

( )Y t vertical position of the unsprung mass

Z global coordinate

Lower Case Times New Roman

a acceleration of the vehicle

na frequency coefficient

c vehicle speed

c viscous damping

0c initial speed of the vehicle

1c primary suspension damping

2c secondary suspension damping

crc critical speed of vehicle

dA infinitesimal area

xviii

dx infinitesimal length in the x-direction

dy infinitesimal length in the y-direction

dz infinitesimal length in the z-direction

e extension

xe unit vector parallel to the local x-axis

ye unit vector parallel to the local y-axis

ze unit vector parallel to the local z-axis

f local nodal force vector of a single finite element

jf natural frequency of the beam in Hz; j = 1, 2 …

( )nf t time function

g gravity

i node number along the beam

mi contact node number using contact elements

1k primary suspension stiffness

2k secondary suspension stiffness

Hk Hertzian spring stiffness

k local stiffness matrix of a single finite element

l local length of an element

rl distance along rigid rail

m mass per unit length

( ),xm x t torque moment about the x-axis

m local mass matrix of a single finite element

j

xp local x components of unit vector due to rotation; j = α, β, γ

xix

j

yp local y components of unit vector due to rotation; j = α, β, γ

j

zp local z components of unit vector due to rotation; j = α, β, γ

( ),p x t pressure force acting over a length x for time t

( ),xp x t axial load parallel to x-axis

( ),yp x t transverse load parallel to y-axis

( ),zp x t transverse load parallel to z-axis

( )tp horizontal and vertical components of the unsprung mass

r radius of a circle measured in radians

( )nr t displacement coefficient

( )nr t& velocity coefficient

( )nr t&& acceleration coefficient

r unit vector

t time variable

jt time of the jth carriage arriving on the beam

u beam displacement in the x-direction

ju beam displacement in x-direction at local node; j = 1, 2

ju&& beam acceleration in x-direction at local node; j = 1, 2

Bju beam displacement in the x-direction at local node; j = 1, 2

( )Bu χ beam displacement in the x-direction at position χ

u local nodal displacement vector of a single finite element

u&& local nodal acceleration vector of a single finite element

v beam displacement in the y-direction

xx

jv beam displacement in y-direction at local node; j = 1, 2

jv&& beam acceleration in y-direction at local node; j = 1, 2

( )Bv χ beam displacement in the y-direction at position χ

Bjv beam displacement in the y-direction at local node; j = 1, 2

w beam displacement in the z-direction

jw beam displacement in z-direction at local node; j = 1, 2

jw&& beam acceleration in z-direction at local node; j = 1, 2

x beam coordinate

1x beam coordinate to the first node on the beam

ix beam coordinate to node i on the beam

Lx beam coordinate to the last node on the beam

y beam coordinate

z beam coordinate

Greek symbols

α dimensionless speed ratio

α beam rotation about the y-axis

0α Raleigh damping constant applied to the global mass matrix

1α Raleigh damping constant applied to the global stiffness matrix

β beam rotation about the z-axis

χ local distance coordinate in the finite element method

δ Dirac Delta function

inδ Kronicor Delta function

xxi

xδθ weighting function about the x-axis

uδ weighting function in the x-direction

vδ weighting function in the y-direction

wδ weighting function in the z-direction

xxε linear strain in the x-direction

xyε shear strain on x-face in the y-direction

xzε shear strain on x-face in the z-direction

( )n xφ n-th characteristic mode shape

( )( )n X tφ characteristic mode shape at the unsprung mass position

( )( )n X tφ′ first derivative of the characteristic mode shape with respect to x

at the unsprung mass position

( )( )n X tφ′′ second derivative of the characteristic mode shape with respect to

x at the unsprung mass position

ϕ phase angle

γ beam rotation about the x-axis

ν Poisson’s ratio

xθ rotation of the beam’s cross-section about the x-axis

xjθ rotational displacement about x-axis at local node; j = 1, 2

xjθ&&

rotational acceleration about x-axis at local node; j = 1, 2

yθ rotation of the beam’s cross-section about the y-axis

ˆyjθ rotational displacement about y-axis at local node; j = 1, 2

ˆyjθ&&

rotational acceleration about y-axis at local node; j = 1, 2

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zθ rotation of the beam’s cross-section about the z-axis

zjθ rotational displacement about z-axis at local node; j = 1, 2

zjθ&&

rotational acceleration about z-axis at local node; j = 1, 2

Bjq beam rotational displacement local node; j = 1, 2

zjθ rotational displacement about z-axis at local node; j = 1, 2

xσ normal stress acting on the x-face in x-direction

yσ normal stress acting on the y-face in y-direction

zσ normal stress acting on the z-face in z-direction

xyτ shear stress acting on the x-face in the y-direction

xzτ shear stress acting on the x-face in the z-direction

yxτ shear stress acting on the y-face in the x-direction

yzτ shear stress acting on the y-face in the z-direction

zxτ shear stress acting on the z-face in the x-direction

zyτ shear stress acting on the z-face in the y-direction

jω frequency of the beam in rad/sec; j = 1, 2 …

bω damping frequency

byω damping frequency parallel to the y-axis

bzω damping frequency parallel to the z-axis

dω frequency of damped oscillation

iξ damping ratio

ψ vector of order n

xxiii

iψ mode shape associated with the eigenvalue

t∆ time taken for a wheel to traverse a beam element

, ,j k m∆ distance between carriages, bogies and wheels

( )n xΦ beam element shape functions; n = 1 to 4

( )( )n X tΦ beam element shape functions at the unsprung mass position

( )( )n X t′Φ first derivative of the beam element shape functions with respect

to x at the unsprung mass position; n = 1 to 4

( )( )n X t′′Φ second derivative of the beam element shape functions with

respect to x at the unsprung mass position; n = 1 to 4

( )( )n X t′′′Φ third derivative of the beam element shape functions with respect

to x at the unsprung mass position; n = 1 to 4

ˆxjΘ global rotational displacement about x-axis at local node; j = 1, 2

ˆxjΘ

&& global rotational acceleration about x-axis at local node; j = 1, 2

ˆyjΘ global rotational displacement about y-axis at local node; j = 1, 2

ˆyjΘ

&& global rotational acceleration about y-axis at local node; j = 1, 2

ˆzjΘ global rotational displacement about z-axis at local node; j = 1, 2

ˆzjΘ

&& global rotational acceleration about z-axis at local node; j = 1, 2

1Ω forcing frequency of the vehicle

Abbreviations

LHS left hand support, Chap 3

RHS right hand support, Chap 3

xxiv

LN1 local node 1, Chap 3

LN2 local node 2, Chap 3

ANSYS CE ANSYS node-to-surface contact element, Chap 3

DAFU dynamic amplification factor of the bridge deflection, Chap 3

DAFA dynamic amplification factor of the bridge axial force, Chap 3

IF impact factor, Chap 3

W1 front wheel of the locomotive, 1st wheel of train, Chap 4

W7 front wheel of first railway coach, 7th wheel of train, Chap 4

V1 node on vehicle body of the locomotive, Chap 4

V2 node on vehicle body of the first railway coach, Chap 4

WRC wheel-rail contact, Chap 4

FEM finite element model used for the unsprung mass, Chap 5

Chapter 1 - Introduction

1

Chapter 1

Introduction

1.1 Background & motivations

For several decades, research on the dynamic response of trains and railway bridges

has become an important topic in civil engineering. Engineers and researchers have a

responsibility for ensuring the safe passage of trains traversing rails and railway

bridges by carrying out extensive research on existing structures. Much of the

dynamic response, which the bridge and vehicle experience, is contributed to by the

contact that exists between the wheel and the rail. Therefore, by modelling the

different wheel-rail conditions, one can better understand the dynamic response of

vehicles and railway bridges.

The different wheel-rail conditions are modelled in finite element programs, such as

ANSYS, by means of contact elements. These contact elements are defined as:

Special finite elements that describe the contact between two bodies, of

which one can move in space, henceforth called the wheel, and the other is

fixed to the ground, henceforth called the rail (Schwab & Meijaard, 2002).


2

Contact elements are used to constrain the vertical, horizontal and lateral positions of

the wheel as it traverses the rail as well as calculate the contact or compressive force

between the wheel and rail. On occasion, wheels of a train can experience a zero

contact force, when there is a loss of contact between the wheel and the rail. In these

circumstances, the position of the wheel relative to the rail is still measured by the

extension of the contact elements.

The original aim of this research was the study of the dynamic interaction of trains

and railway-bridges, where the ANSYS contact element was to be used to model

wheel-rail interactions. The Boyne Viaduct is the particular bridge of interest to the

author. However, it was discovered that the ANSYS contact element was unable to

model wheel and rail irregularities and braking. Therefore, the author developed a

wheel-rail contact (WRC) element which resulted in three stiffness matrices. The

author could input these matrices to the ANSYS program. This development is

discussed in greater detail in the following chapters and allows the author to

investigate the following rail scenarios:

• Smooth-rail condition

• Wheel & rail irregularities

• Wheels travelling on a rigid rail

• Wheel losing contact with rail

• Braking & accelerating effects

• Lateral effects on the vehicle

The author was able to address the list of scenarios above at the annual UK ANSYS

User Conference in Warwick in 2002 (Bowe & Mullarkey, 2002).


3

The numerical models developed in this study are validated using the results from the

literature. In many cases, the bridges consist of beams. The Boyne Viaduct railway-

bridge, located on the Dublin-to-Belfast line in Drogheda, provides a complex three-

dimensional structure on which to test the author’s models.

Besides the WRC model, this thesis also models a moving unsprung mass traversing a

beam or rail. This differs from the moving sprung mass where the spring is simulated

by either the ANSYS contact elements or the WRC element. The unsprung mass has

no spring. The vertical acceleration of the unsprung mass is the sum of the vertical

acceleration of the bridge and a convective acceleration. In the Biggs (1964) and Akin

& Mofid (1989) moving unsprung model, the convective acceleration is omitted from

the model; thus their solution is inaccurate. Nevertheless, many authors are still

comparing the results of their models to this inaccurate solution and are ignoring

convective acceleration. This thesis derives both a modal and finite element model

incorporating a moving unsprung mass and an irregular rail.

1.2 ANSYS Strengths & Weaknesses

In the initial stages of this study it seemed very advantageous for the author to use the

ANSYS finite element program to model the bridge-train systems, as this is a multi-

purpose finite element program that is used worldwide. The author required a finite

element program with static, modal and transient analysis capabilities so that one

could simulate the effects of trains traversing rails and railway bridges. Some of the

reasons for choosing ANSYS the finite element program were:


4

• Element Library – The program has a large library of structural elements that

would be required by the author. Two-dimensional and three-dimensional

elastic beam elements [BEAM3 & BEAM4] could model beams, rail and

vehicle components, while lumped mass elements [MASS21] would be used

to simulate the wheels of the vehicle. Suspension springs between the wheels

and the vehicle components would be represented by spring damper elements

[COMBIN14], whereas node-to-surface contact elements [CONTAC48]

would be used to model the wheel-rail interaction. Another element that

proved invaluable later in this study is the matrix element [MATRIX27]. This

is an empty 12x12 mass, stiffness or damping matrix created for any pair of

nodes and has been utilized extensively throughout to simulate the WRC

element as well as the modal and the finite element models incorporating the

moving unsprung mass.

• Equation Solvers – In the solution phase of the analysis, equation solvers are

used to solve the set of simultaneous equations that are generated in the finite

element model. For transient analysis one uses the linear solver Newmark-β

for time integration.

• Result as a function of time – Since the majority of the author’s studies

involve transient analyses, one requires the results to be plotted as a function

of time. These results would include deflections, contact forces, axial forces

and bending moments.

• APDL Code – ANSYS Parametric Design Language or APDL is a scripting

language that allows the author to automate common tasks and build models in


5

terms of parameters. Mastering the APDL code can be tedious, but the rewards

are immense. Like other computer languages it has a wide range of features

such as macros, conditional branching, do-loops, and scalar, vector and matrix

operations.

Thus far one has discussed advantages of ANSYS. However, ANSYS has many

limiting features, which the author eventually overcame. ANSYS limitations include:

• No point force between nodes – In the early stages of the research, moving

point forces were use to model the train wheels traversing a beam or rail.

However, the author discovered that ANSYS can only apply point forces to

nodes along the beam. Point forces could not be applied between the nodes.

The author developed a system whereby forces and moments are applied to the

two nodes of a beam element as a function of time to represent the point force

travelling between these two nodes.

• Contact element cannot model irregularities – Both the two-dimensional

and three-dimensional node-to-surface contact elements [CONTAC48 &

CONTAC49] are unable to simulate wheel or rail irregularities. These

elements can only model smooth-rail conditions. Fortunately with the aid of

the empty 12x12 stiffness matrix [MATRIX27] in ANSYS, the author was

able to develop a wheel-rail contact element incorporating wheel and rail

irregularities.


6

• Three-dimensional contacts required a surface area – Another drawback of

using the three-dimensional node-to-surface contact elements [CONTAC49]

was that one has to model the rail as a solid block element instead of as a

beam element. In addition, CONTACT49 neglects the lateral support of the

wheels. Using the author’s wheel-rail contact element, one was able to model

the vertical, longitudinal and lateral effects of the wheel on a three-

dimensional beam element.

• Real constant cannot be specified as a function of time – Since the real

constant values, i.e. section properties of the element, cannot be specified as a

function of time within ANSYS, the author has to update these constraints by

stopping and starting the ANSYS program. The real constant values in the

matrix elements [MATRIX27] need to be updated at every timestep. The real

constant values in the matrix elements change when the position of the wheel

changes.

1.3 Chapter Summary

Chapter 2 consists of a discussion of other researchers’ work related to the author’s

study. The chapter is designed to give the reader some insight to this project by

examining different approaches to modelling wheel-rail interaction and how they

differ from the author’s systems. One also examines other aspects of this study, such

as the dynamic response of the vehicle and railway-bridge.

A procedure that represents the wheels of a train as a series of moving forces is

presented in Chapter 3. Forces are applied to nodes along the bridge or beam as a


7

function of time to simulate the passage of the train. This system was developed

because the author could not apply point forces between nodes in ANSYS. Originally

the technique was setup so that a beam finite element never had to support more than

one point force. Thus a large number of closely spaced wheels of a train would

require a large number of bridge elements and nodes. This increased the execution

time of the analysis. Later, one was able to allow one element to support several point

forces using APDL code. This dramatically reduced the number of nodes in the model

and resulted in a quicker execution time. Moving force models are generally

allowable if the mass of the vehicle is substantially less than the mass of the bridge.

One can then ignore the inertia effects.

The wheel-rail contact (WRC) element is developed by the author in Chapter 4. This

system involves modelling each wheel as a point mass at the centre of the wheel. The

contact element is then placed between the point mass and the rail. The contact

element uses three stiffness matrices to model the contact. The first stiffness matrix is

for the two nodes of the beam element, the second stiffness matrix is for the first local

node of the beam element and the centre node of the wheel and the third stiffness

matrix is for the second local node of the beam element and the centre node of the

wheel. The non-zero values inputted into these stiffness matrices are a combination of

the contact spring stiffness and beam element shape functions, evaluated at the

wheel’s position. The author compares and contrasts the results using the spring

element contact with the results using the node-to-surface contact elements

[CONTAC48] within ANSYS in order to validate the model. Published results and

analytical results are also used for comparison purposes. The author also develops

both the longitudinal and lateral spring elements to support the wheel longitudinally


8

and laterally. Thus accelerating and braking forces as well as traverse loadings are

possible. The lateral spring elements provide lateral support to the wheels of the

vehicle in three-dimensional analyses.

In Chapter 5 the author examines an alternative to the sprung mass wheel system,

namely the unsprung mass wheel system. The wheel is assumed to be permanently

attached to the beam and cannot separate from it. Thus both the local and convective

accelerations of the unsprung mass must be taken into account. In this chapter, the

author develops both a modal and finite element model incorporating the moving

usprung mass. Both systems have many similarities. The modal method uses modal

characteristic shapes and displacement coefficients, while the finite element solution

uses element shape functions and displacement values at nodes to represent

displacements at all points of the beam. While computing the convective acceleration

terms for the unsprung mass, one must take second derivatives. In the finite element

solution, the second derivative of the element shape function becomes discontinuous

at the nodes. In the modal method, there are no nodes and the second derivative of the

modal characteristic shapes remains continuous at all times. The difference between

the modal and finite element model occurs in the representation of the stiffness

matrix. Integration by parts gives the finite element stiffness matrix a different form to

the modal stiffness matrix

Until now one has considered smooth rail conditions. In Chapter 6, the sprung and

unsprung systems are extended to include the effects of irregularities along the rail.

The author uses a deterministic approach for modelling the irregularity function.

Irregularities on the rail can lead to wheels of a vehicle losing contact with the rail.


9

This can occur when the extension in the wheel-rail contact (WRC) element for a

particular wheel becomes positive, and all the values in the stiffness matrices related

to that wheel become zero. As a train traverses the rails, the railway carriage can also

experience pitching motion if irregularities on both rails are the same and in-phase or

rolling motion when irregularities on the rails are out-of-phase. This can lead to a

reduction of the passenger riding comfort. In addition, one also modifies the moving

unsprung mass to incorporate the effects of irregularities for both the modal and finite

element systems. The final chapter summarizes the results given in this study as well

as making recommendations for further research in this area. The key themes

examined by the author in this thesis can be seen in Figure 1.1.

The reader should also refer to the Appendices for all technical data and structural

theories discussed in this study. In particular, Appendix A establishes a consistent

sign convention; it develops moment-curvature relationships as well as deriving the

Bernoulli-Euler differential equation for an elastic beam and spring, while Appendix

B then applies the finite element method to the developed differential equations for

the beam and spring. The natural frequencies and modal shape functions for a

cantilever beam, encastre beam and simply supported beam are then derived in

Appendix C, while the equation of damping and Newmark time integration scheme

are derived in Appendix D and E, respectively. Technical bridge parameters related to

the Boyne Viaduct are given in Appendix F, while railway vehicle dynamics related

to the author’s vehicle model is given in Appendix G.

Chap

ter 1 - In

trod

uctio

n

10

Hertzian springbottom of

No Seperationstiffness equal to infinite using MATRIX27

Seperation

Lateral3D structures

Longitudinalbraking effects

implementation

Constraint

Horizontal

including irregularities

Contact

HertzianStiffnessStiffness

Non-Hertzian

WheelInertia

using MASS21

WheelInertia

local & convective acceleration

Vehicleincluding railway carriages

WheelSingle

WheelsMultiple

Forceweight of wheel

Beam

using BEAM4

FEM

using MATRIX27

ModesAnalytical

2D Truss3D Boyne

centre ofthe wheel

Chap 3

Chap 4

All Chap All Chap

Chap 4

using Newmark-B

Transient

Chap 5 Chap 4

All ChapAll ChapAll Chap

Chap 4Chap 5

Chap 4Chap 5

Chap 4Chap 5All Chap

All Chap

All Chap

Chap 5

Fig

ure 1

.1

Key th

emes exa

min

ed in

this th

esis

Chapter 2 – Literature Review

11

Chapter 2

Literature Review

2.1 Introduction

The dynamic interaction of trains and railway bridges has several components. These

components are the vehicle and its characteristics; the bridge and its characteristics;

the wheel-rail interaction including the effects of wheel and rail irregularities. In this

chapter, the author discusses each component separately using the literature. Varying

certain parameters of each component can have a significant effect on the overall

outcome of the problem, e.g. increasing the vehicle speed generally increases the

maximum dynamic deflection of bridge.

2.2 Modelling Railway Vehicles

2.2.1 History of Railway Vehicle Dynamics

According to Esveld (2001), an Englishman by the name of Trevithick constructed the

first ever steam locomotive in the early nineteenth century. However, it would take

several decades, and a railway bridge disaster, before any theoretical or experimental

studies of railway bridges dynamics would commence. In 1847 following the collapse

of the Chester Railway Bridge in England two pioneers emerged to study railway


12

bridges (Yang et al., 2004). These early pioneers were Willis and Stokes. Willis

solved the problem of a moving load crossing a simply supported beam such that the

mass of the beam is substantially less than the mass of the vehicle, while Stokes was

first to obtain experimental data related to this problem (Fryba, 1999).

Not until the early twentieth century did Krylov, Zimmermann and Timoshenko begin

studying the problem of a single concentrated load crossing a simply supported beam

(Fryba 1996). Their studies assumed that the mass of the vehicle was substantially

less than the mass of the beam and ignored vehicle inertia. However, Akin & Mofid

(1989) pointed out that, by ignoring the inertia of the vehicle whose mass is about the

same as the mass of the bridge, there could be an error of 20 to 80% in the dynamic

deflection of the bridge. The error depends on the mass velocity and boundary

conditions of the beam, although their model only took into account the local

accelerations of a moving unsprung mass. In 1921, Saller was first to study both the

gravitational and inertial effects of the moving vehicle crossing a bridge structure. His

work was followed by Jeffcott in 1929, by Steuding in 1934 and by Odman in 1948,

according to Fryba (1999).

The classical study of pulsating or harmonic loads traversing a simply supported beam

at a constant speed is credited to Timoshenko in 1922, while Inglis provided a much

more detailed study in 1934. This was to represent the harmonic action of unbalanced

counterweights of the driving wheels of steam locomotives (Fryba, 1996). Inglis

(1934) examined the dynamics of a steam locomotive traversing a railway bridge both

theoretically and experimentally. This included the motion of the bridge subjected to a

concentrated load, sprung and unsprung masses and a harmonic force crossing a

beam.


13

Fryba (1999) tells that Hillerborg was first to solve the problem of a motion of a

sprung mass crossing a simple beam by means of Fourier’s method and the method of

numerical difference in 1951. With the introduction of digital computers, further

advances in this topic were made possible. Using Hillerborg’s method, Tung et al.

(1956) solved the vibration of highway bridges, while Biggs et al. (1959) utilized

Inglis’s method.

Apart from the initial fundamental models, there have been many studies of railway

vehicle dynamics by recognised researchers (Biggs, 1964; Chu et al., 1979; Garg &

Dukkipati, 1984, Esveld, 1989; Akin & Mofid, 1989; Yang & Yau, 1997; Fryba,

1999, Esveld, 2001; Iwnicki, 2006). These authors have modelled realistic bridges

and trains, where the trains can travel at high speed or at variable speeds.

2.2.2 Vehicle Model

This section briefly discusses the different types of vehicle models used by the author

in this thesis to simulate a train traversing a railway bridge.

2.2.2.1 Moving Constant Force

Perhaps the most common and simplest vehicle model comprises several moving

constant point forces i.e. only the weight of the train is taken into account. This type

of vehicle model is generally associated with railway bridges of medium or large

spans (> 30m) because the mass of the bridge tends to be substantially greater than the

mass of the vehicle.


14

When it comes to solving this problem, there are many different solutions available.

Fryba (1999) solves the problem using a Fourier integral transformation of the

equation of motion of the beam. The same procedure is also adopted by Green &

Cebon (1994), Yang et al. (1997) and Wang et al. (2003). Biggs (1964) uses a series

of beam modes to calculate the deflection of the beam due to the moving force. This

amounts to Fryba’s solution. A finite element procedure whereby linear forces are

applied to every node along a beam as a function of time to represent the moving

constant force is presented by Bowe & Mullarkey (2000). A study by Wu et al. (2000)

represents the moving constant force as forces and moments applied to the nodes of

the beam as a function of time. In Yang et al. (1997) as well as Yau & Yang (2006)

and Garinei & Risitano (2008), the timing issue of several moving loads arriving on a

beam are handled by Heaviside functions.

2.2.2.2 Weight and Mass of Moving Wheel

There is greater complexity when both the gravitational and inertia effects are

included. Throughout the literature, the wheel mass can be modelled as a sprung or

unsprung mass. Some authors (Akin & Mofid, 1989; Biggs, 1964; Chatterjee et al.,

1994; Lee 1998; Yang et al., 1997; Yang & Wu, 2001) model the wheel as an

unsprung mass always in direct contact with the beam. Thus the deflection of the

wheel and beam are equal at the point of contact. Other authors (Esveld, 1989; Zhai

& Cai, 1997; Zhang et al., 2001; Sun & Dhanaekar, 2002; Iwnicki, 2006; Liu et al.,

2008) place a Hertzian spring between the wheel and the beam. Hence, the deflection

of the wheel and beam differ at the point of contact. Since the wheel and rail share a

common contact point in the unsprung system, convective velocity and acceleration

terms are needed for the wheel.


15

This is not applicable to the sprung system. Fryba (1996) adds convective terms but

many authors such as Biggs (1964), Akin & Mofid (1989) and Yang et al. (2004) do

not. Unlike the unsprung mass, wheels that are separated from the rail by a Hertzian

spring can separate from the rail when the spring becomes tensile. Such separation is

studied by Zhai & Cai (1997), Bowe and Mullarkey (2002), Sun & Dhanaekar (2002)

and Liu (2008).

2.2.2.3 Mass of the Moving Vehicle

The moving vehicle comprises wheels supporting a sprung vehicle by means of a

primary suspension. The suspension often consists of a spring dashpot as illustrated

by Biggs (1964), Chang & Liu (1996), Rieker et al. (1996), Green & Cebon (1997),

Cheng et al. (2001), Yang & Wu (2001) and Ju & Lin (2003). The force applied to the

beam by the moving vehicle is due to the gravitational and inertia effects of the

wheels and vehicle.

A typical railway vehicle can also consist of a vehicle body supported by a pair of

bogies, with each bogie supported by two axles and finally a pair of wheels supports

each axle. The bogies are connected to the axles and to the railway vehicle body,

respectively, through the primary and secondary suspension systems, with each

suspension consisting of a spring and dashpot similar to that presented by Garg &

Dukkipati (1984), Xia et al. (2000) and Esveld (2001). In Ireland, the locomotives

tend to be supported by twin bogies, with three axles supporting each bogie, while in

rest of Europe and around the world the number of axles supporting a locomotive can

vary.


16

Using the works of Garg & Dukkipati (1984), Xia et al. (2000) and Zhang et al.

(2008), one can describe the motion experienced by a three-dimensional railway

vehicle as it travels. Each wheelset of a vehicle has three degrees-of-freedom

(vertical, lateral and rolling motion), while each bogie and vehicle body has five

degrees-of-freedom (vertical, lateral, rolling, yawing and pitching motion), where the

rolling, yawing and pitching motion are defined as rotations about the x-axis, y-axis

and z-axis, respectively. Xia et al. (2000) models a three-dimensional, four-axle

railway vehicle using 27 degrees-of-freedom, while a similar two-dimensional railway

vehicle given by Au et al. (2002) is described by only 10 degrees-of-freedom. It

should be noted that in the two-dimensional model, only the vertical and pitching

motions of the vehicle are taken into account. Studies conducted by Zhang et al.

(2001) and Sun et al. (2003) have also examined the lateral motion of a railway

vehicle as it travels, while Yang et al. (1999) investigates the pitching motion of vehicles.

2.2.3 Vehicle Characteristics

Using the literature, one now discusses the different vehicle characteristic that can

greatly influence the outcome of the problem.

2.2.3.1 Vehicle Spacing

The importance of vehicle spacing is best illustrated in Yang et al. (1997), Fryba

(2001) and Yau & Yang. (2006). According to Fryba (2001), closely spaced axles or

vehicles can cause the resonant vibration of railway bridges, even at low speeds. The

speed, at which resonance is likely to occur on the bridge can be defined as the first

natural frequency of the bridge multiplied by the repetitive distance between axles,

bogies or railway vehicles traversing the bridge. According to Fryba (2001), medium


17

to large spanning bridges, with lower natural frequencies, that are subjected to short

railway vehicles are more susceptible to experiencing resonance vibration e.g. a 40 m

bridge with its first natural frequency equal to 2.5 Hz would resonate at a vehicle

speed of 27.8 km/hr if the repetitive distance between vehicles is equal to 3 m.

However, Li and Su (1999) states that if the number of vehicles traversing the bridge

is very small, resonances may not occur. Yang et al. (1997) observed that a condition

of cancellation that suppresses the resonant vibration of the bridge can occur when the

ratio of the bridge length to the repetitive distance of carriages coincides with the

following values 0.5, 1.5, 2.5, 3.5 etc, for a simply supported beam.

2.2.3.2 Vehicle speed

The next characteristic that one considers is the speed at which the vehicle travels. A

vehicle travelling along a bridge at a low speed behaves in a similar manner to a

stationary vehicle; thus the dynamic deflection experienced by the bridge can be low.

A vehicle travelling at a high speed can significantly increase the dynamic deflection

of the bridge. The maximum vibration of railway bridges can occur if certain critical

speeds are reached. Fryba (1999) calculates the critical speed of the vehicle as the

speed of the vehicle such that the vehicle travels a distance of twice the length of the

bridge in a time equal to the natural period of the bridge.

In Fryba (2001) vehicles travelling at a wide range of critical speeds across different

bridge lengths are studied. It is found from the study that the critical speeds often

exceed the speed even of high-speed vehicles e.g. a 40 m bridge with its first natural

frequency equal to 2.5 Hz would resonate at a vehicle speed of 720 km/hr.


18

Olsson (1985), Cheung et al. (1999) and Fryba (1999) have examined the effects of a

moving load, sprung mass and unsprung mass traversing a railway bridge at different

speeds. It is observed that the dynamic response of the bridge tends to increase with

the increasing vehicle speed. However, observations by Fryba (1999) have shown that

the dynamic response of the bridge can also decrease with increasing speed especially

at high speeds greater than the critical speed. Cai et al. (1994) finds that the maximum

deflection of a bridge occurs at less than 0.5 the critical speed for a moving force

model whereas the moving sprung mass system tends to have a maximum deflection

closer to a critical speed of 1. The resonant vibration of simply supported bridges

under high-speed trains has also been investigated by other authors such as Li and Su

(1999) and Xia & Zhang (2005). Findings from Xia & Zhang (2005) have shown that

derail factors and lateral wheel-rail forces of the vehicle also increase with the train

speeds.

2.2.3.3 Braking and accelerating of vehicles

When studying railway dynamics, most authors seem to be more concerned with the

vertical response of the bridge and vehicle; thus the vehicle traverses the bridge at a

constant speed. The horizontal response of the bridge and vehicle due to braking and

accelerating is often ignored. However, according to Esveld (2001) the horizontal load

resulting from a vehicle braking can often be as much as 25% of the vehicle’s weight,

found by multiplying the vehicle weight by the coefficient of friction of the vehicle.

In Fryba (1999), three different cases of a moving load traversing a bridge at variable

speeds are considered. Firstly, a vehicle travelling at a constant speed begins to

decelerate uniformly at the instant it arrives on the bridge (x0 = 0) such that it comes


19

to a complete stop at the end of the bridge (x = l). Secondly, a vehicle at rest (x0 = 0)

begins to accelerate uniformly such that it has a constant speed the instant it arrives at

the end of the bridge (x = l). Thirdly, A vehicle travelling at a constant speed begins to

decelerate uniformly at the instant it arrives on the bridge (x0 = 0) such that it comes

to a complete stop at the mid-span of the bridge (x = l/2). Fryba (1999) finds that the

uniform deceleration of the moving load crossing the bridge in the 1st and 3

rd example

results in a higher dynamic deflection than that of a uniform acceleration in the 2nd

example. In addition, Fryba (1999) remarks that there is little difference in the results

between 1st and 3

rd example even though the acceleration of the latter is twice the

former.

In Bowe & Mullarkey (2002), the effects of a locomotive braking along a railway

bridge are examined. The instant the front wheel of the locomotive enters a bridge,

travelling at a constant speed, it begins to decelerate uniformly such that the

locomotive comes to a complete stop when its front wheel reaches the end of the

bridge. As the wheels enter the bridge, it is shown that there is a significant increase

in the horizontal force experienced by the bridge support. Then as the locomotive

slows to a stop there is sudden drop in the force and finally it oscillates about its

equilibrium state. These observations were also made by Yang & Wu (2001).

Vehicles subjected to a forward or retard force in accordance with Newton’s 2nd

law

of motion can be used to model the acceleration and deceleration effects of a vehicle.

In Wang (1998), forward and retard forces are used to accelerate and decelerate a

moving mass traversing a simply supported beam.


20

2.2.3.4 Passenger Riding Comfort

In the study of railway vehicle dynamics, one is particularly interested in the

vibrations of the vehicle body as it travels, as these vibrations affect passenger riding

comfort. According to Yau et al. (1999), the vertical and lateral accelerations of the

vehicle body serve as a measure of the riding comfort. The maximum allowable

vertical acceleration adopted by France-SNCF is 0.49m/s2, while the Eurocode (1990)

has implemented a less stringent range of values. The level of passenger riding

comfort ranges from very good to acceptable in the Eurocode (1990). For the riding

comfort to be considered very good, the vertical acceleration must have a value less

than 1.0 m/s2, while a value less than 2.0 m/s

2 is deemed an acceptable. It is observed

in Yau et al. (1999) and Wu & Yang (2003) that the presence of track irregularity can

greatly affect the passenger riding comfort of the train, especially at medium to high

speeds.

A more detailed study into passenger riding comfort is presented in Garg & Dukkipati

(1984). These authors compute the ride quality and the comfort of the rail vehicle by

means of a ride factor, which comprises the vehicle acceleration amplitude and an

acceleration-weighting factor. According to Garg & Dukkipati (1984), a ride factor of

less than 1 gives a very good ride quality and a barely noticeable vibration comfort,

while a ride factor of greater than 4 provides an unacceptable ride quality and an

extremely unpleasant vibration comfort.

2.2.3.5 Ratio of between sprung vehicle mass and unsprung wheel mass

According to Fryba (1999), the ratio between the sprung vehicle mass and unsprung

wheel mass for a single moving sprung mass system tends to have little effect on the

dynamic amplification factor of bridges.


21

2.2.3.6 Ratio of the vehicle mass to bridge mass

The relationship between the vehicle mass P and bridge mass G becomes important in

structural dynamics when the vehicle mass is comparable to or greater than the mass

of the bridge, especially at high speeds (Inbanathan & Weiland, 1987). Fryba (1999)

demonstrates that as the vehicle to bridge mass ratio grows the dynamic response of

the bridge also increases. It should also be noted that as the P/G ratio becomes

significantly small, i.e. less than 0.2, the inertia effects of the vehicle are reduced;

hence, there is little difference between modelling the vehicle as a load and modelling

it as a mass traversing a beam as described in Gbadeyan & Oni (1995) and Ickikawa

et al. (2000).

2.3 Modelling Railway Bridges

This section focuses on the different railway bridge types used in this study and it also

discusses the bridge characteristics and how they can influence the dynamic behaviour

of the structure. The author is particularly interested in modelling the centre span of

the Boyne Railway Viaduct as a case study. This comprises a steel truss railway

bridge with a clear length of 80.77m between supports.

2.3.1 Railway Bridge Types

The main types of bridge structures that one focuses on in this study are beam and

truss railway bridges, while all other bridge types are less important.

2.3.1.1 Beam Railway Bridges

Perhaps the simplest and most commonly studied bridge type is a beam railway

bridge. Due to its simplicity many researchers, such as Cheung et al. (1999), Delgado


22

& Dos Santos (1997), Yang et al. (1997) and Bowe and Mullarkey (2005), use a

simply supported or cantilever beam model to validate their works before modelling

more sophisticated bridge structures. In these studies, the beam generally behaves in a

linear-elastic manner, with a constant beam mass per unit length. The model may

include damping. Fryba (1999) represents a truss railway bridge as a simply supported

beam using some simple calculations. Firstly, one calculates the mass of the beam per

unit length by dividing the total weight of the bridge structure G by the acceleration

due to gravity as well as the bridge length, while the cross-section areas of the upper

and lower chords of the truss at the appropriate truss height are used to compute a

suitable moment of inertia for the beam. Alternatively, if one knows the maximum

static deflection of the truss due to its own self-weight, one can calculate a suitable

moment of inertia for the beam using the beam deflection equation of a uniformly

distributed load.

2.3.1.2 Truss Railway Bridges

In the early days of railway bridge construction, steel riveted trusses were used

extensively on medium to large spanning structures. In modern times, with the

introduction of new construction methods and materials, the steel riveted truss railway

bridge has been superseded by bolted and welded structures. Nonetheless, many

existing railway bridge are steel riveted truss structures. Studies involving the

dynamic interaction of trains and steel truss bridges can be found in Chu et al. (1979),

Wiriyahai et al (1982), Garg & Dukkipati (1984), Wang et al. (1991), Delgado & Dos

Santos (1997) and Ermopoulos and Spyrakos (2006).

Presently, researchers examining steel riveted railway trusses are studying fatigue.

This is not an objective of the author’s project.


23

2.3.1.3 Suspension & Cable Stayed Bridges

Xia et al. (2000), Au et al. (2002) and Yau & Yang (2004) have carried out extensive

studies of trains traversing suspension and cable-stayed bridges; however, these types

of bridge structure are not included in the present study.

2.3.1.4 Boyne Viaduct Railway Bridge

As a case study, the Boyne Viaduct Railway Bridge located in Drogheda is examined,

which is located on the Dublin to Belfast railway line. Figure 2.1 presents the bridge

structure, which consists of two outer trusses and a central-curved truss supported by

masonry piers. A detailed description and historical assessment of the bridge can be

found in Howden (1930). The reader should refer to Appendix F for a more detailed

account of this bridge structure.

Figure 2.1: Boyne Viaduct Railway Bridge

A structural assessment of the Boyne Viaduct railway bridge subjected to a moving

locomotive is presented in Gallagher (2002). In this study, Gallagher uses a

FORTRAN based program to analyse the bridge-train system. Despite certain results

from this study proving to be inconclusive with other research literature, one can get a


24

good insight into the behaviour of the Boyne Viaduct railway bridge as vehicles

traverse it. In Majka et al. (2002), the Boyne Viaduct railway bridge is subjected to a

sprung vehicle with different Hertzian stiffness. It is found from this particular study

that varying the Hertzian stiffness has little effect on the bridge but significant effect

on the vehicle, especially for the riding comfort felt by the passengers.

2.3.2 Bridge Characteristics

Using the literature, one now discusses the different bridge characteristics that can

greatly influence the outcome of the problem.

2.3.2.1 Bridge Natural Frequency

Perhaps one of the most important dynamic characteristics in the study of railway

bridges is the natural frequency, as this indicates at which frequencies the bridge is

sensitive to dynamic loads. The unit of frequency is the Hertz (Hz), which is the

number of cycles or vibrations per second. The natural frequency of the bridge is a

function of its length, mass and bridge stiffness. In Appendix C, the author defines the

natural frequency for a simply supported, cantilever and fixed-fixed beam.

Alternatively, Fryba (1996) describes several empirical formulae to calculate the first

natural frequency of the bridge structure. Firstly, the natural frequency can be

computed empirically by dividing a value of 5.62 by the static deflection of the bridge

under its own self-weight, measured in centimetres. Secondly, one divides the value

of 250 by the length of the bridge. In addition, Fryba (1999) plots the first natural

frequency of steel railway bridges as a function of its span length. It is found that a

bridge with a length of 80.77 m has a natural frequency approximately equal to 3.2

Hz. Finally, the natural period of vibration T is computed as the inverse of the natural


25

frequency. In dynamics, it is recommended that the timesteps used in the integration

be a fraction of the shortest natural period of vibration being examined (usually a

tenth), so that natural frequencies in the model are not overlooked.

2.3.2.2 Bridge Damping

In structural dynamics, damping plays a vital role in reducing the dynamic response of

the structure as well as in causing a bridge to return to its state of equilibrium soon

after a vehicle has passed (Fryba, 1996). Bridges can have many sources of damping

ranging from the internal friction of the construction materials to the friction of

support bearings or joints. The assumption of viscous damping, used by many

researchers, is where the damping is proportional to the velocity of vibration of the

bridge (Fryba, 1999).

From an extensive study of railway bridges, Fryba (1996) found that there was very

little correlation between damping and the span, the bending stiffness, the deflection

due to the self-weight of the bridge, the dynamic stiffness and the first natural

frequency of the bridge. It was also noted that damping had little or no effect on

bridges with and without ballast.

Fryba (1996) uses several empirical formulae to compute the damping of the bridge

structure. In one such empirical formula, the damping is calculated using the length of

the bridge, while in another formula the natural frequency of the bridge is used.

In Yau et al. (2001), the effects of damping are examined on the resonance response

of a beam subjected to 30 moving loads. It is found from their study that the vibration


26

of a damped beam seems to remain in a stable manner, even under the conditions of

resonances, while the amplitude of the undamped beam tends to grow increasingly as

more loads traverse the beam. Kwon et al. (1998) present a paper on the vibration

control of bridges under moving loads using a tuned mass damper (TMD) installed in

the middle of the bridge. In their particular study, it is found that the maximum

displacement induced by a high speed train is decreased by 21% and free vibration

dies out rapidly with the introduction of the tuned mass damper.

When computing a suitable structural damping, many researchers such as Wu & Yang

(2003), and Zhang et al. (2008) tend to use the Rayleigh damping scheme, which is a

linear combination of the mass matrix and stiffness matrix multiplied by coefficients;

whereby these coefficients are a function of the damping ratio and the natural

frequencies of the bridge.

2.3.2.3 Bridge Span Length

As can be seen from the previous two subsections, the bridge span length L governs

the natural frequency and damping coefficient of the bridge structure; hence,

increasing or decreasing the span length can dramatically affect the dynamics. One

should also be aware that short spanning bridges behave differently from longer

spanning bridges when vehicles traverse them. For example, short bridges subjected

to long trains may only experience the action of a single bogie on the structure at any

one time; hence, as a bogie leaves the bridge, the bridge will begin to vibrate freely.

Subsequently, the following bogie arriving on the bridge will experience the

vibrations. Delgado & Dos Santos (1997) concluded that as the length of span

decreases the dynamic response of the bridge structure increases with increasing

speed, especially at high speeds. Yang et al. (1997) made a similar observation.


27

2.3.2.4 Rail approaches

Rail approaches located on either side of a bridge allow long vehicles to enter and

leave the bridge structure. The rail approaches are generally modelled as rigid rails,

whereas the bridge structure is flexible. Chu et al. (1985), Bogacz & Kowalska (2001)

and Bowe & Mullarkey (2005) have adopted rigid rail approaches in their bridge

models.

2.3.2.5 Dynamic amplification factor

The dynamic amplification factor can be defined as the ratio of the maximum

dynamic response to the static response of a bridge. This response is often measured

by the deflection of the bridge, but can also be measured by the internal forces of the

structure. Throughout the literature, the dynamic amplification factor has many name

variations.

Fryba (1999) describes it as the dynamic coefficient, while Au et al. (2002) refers to it

as an impact factor, Museros et al. (2002) calls it an impact coefficient, whereas

Cheng et al (2001) describes it as a dynamic magnification factor. Nonetheless, in all

cases it remains the ratio between the dynamic and static response. In this study, one

will refer to this term as the dynamic amplification factor or DAF.

A comprehensive study of the dynamic amplification factor for a train traversing a

railway bridge at different speeds is presented in Delgado and Dos Santo (1997). The

dynamic amplification factor is influenced by the damping on the structure; the train

stiffness; the ballast on the bridge; rail irregularities and the bridge length. Their study

found that the bridge stiffness and varying the bridge length were most influential.

Flexible bridge structures give high amplifications, while rigid bridge structures give


28

lower amplifications. In addition, track roughness can increase the dynamic

amplification factor of the bridge, especially if the irregularity coincided with the

natural period of the bridge or train.

In Figure 2.2, the author presents the dynamic amplification factor observed by many

researchers (Cai et al., 1994; Cheung et al., 1999; Fryba, 1999; Savin, 2001; Wu &

Yang, 2003; Majka & Hartnett, 2008) for the deflection of a bridge plotted as a

function of vehicle speeds with bridge damping giving the reader a family of curves.

From the diagram, one can see that the critical speed parameter of the vehicle occurs

at α = 0.6, while the critical damping parameter occurs at β = 1. As found by most

researchers, the DAF tends to increase with increasing speed, but can also decrease at

very high-speeds or with damping included.

Figure 2.2: DAF of bridge deflection subjected to different speeds (Fryba, 1999)

2.3.3 Numerical Bridge Solutions

When it comes to analysing the dynamic behaviour of bridge structures, researchers

have commonly used the following two techniques; the finite element method and the

method of modal superposition. The similarities and differences between the modal

and finite element model, as far as the form of the final matrices are discussed in

Dyn

am

ic a

mp

lifi

cati

on f

act

or


29

Bowe & Mullarkey (2008). In addition, the weighting and shape functions of the

finite element method play a similar role as the mode shapes play in the modal

solution. The difference between the two methods occurs in the representation of the

stiffness matrix.

2.3.3.1 Finite Element Method

The finite element method is a technique for analysing complicated structures by

theoretically cutting up the continuum of the structure into a number of small

elements that are connected at discrete joints called nodes (Hambly, 1991). For each

element, the approximate stiffness equation is derived by relating its nodal

displacements with the nodal force on that element. Computers are then used to solve,

often large numbers of, simultaneous equations that relate the nodal forces and

displacements for the entire structure. The finite element method is an approximation

of a real structure and the construction of the finite element models can often affect

the accuracy of the results. Generally, the larger the number of elements used to

discretize the beam or structure, the more accurate the solution (Fenner, 1996).

The finite element method can also be described as a numerical means for finding the

approximate solutions of partial differential equations. As illustrated in Appendix A

as well as in Bathe (1996) and Hutton (2004), the vibration of an elastic beam

subjected to a transverse load can be written in the form of a partial differential

equation. The beam equation contains a fourth order bending stiffness term plus a

second order mass term for the beam. In order to solve this partial differential

equation, both terms are integrated twice; thus, been replaced by weighting function

and shape functions using a technique known as the Galerkin Method, which is


30

presented in Appendix B. This technique is also described in most finite element

books by authors such as Bathe (1996), Zienkiewicz (2000), Hutton (2004) and Fish

& Belytschko (2007). ANSYS is the finite element program used by the author

throughout this thesis, and is used to solve both the finite element models as well as

modal superposition models.

2.3.3.2 Method of Modal Superposition

The method of modal superposition is a technique that can be used for analysing

simple structures such as simply supported beams, cantilever beams and continuous

beams subjected to a moving load or vehicle. Modal superposition treats each mode of

vibration of the entire beam as a single degree of freedom oscillator and then

combines the responses of these individual modes into a complete dynamic solution.

When using the method of modal superposition, it is important that a sufficient

number of modes be used to properly capture the dynamic deflection of the beam or

structure. In order to capture the internal forces of the beam, one requires several

more modes of vibration than the deflection of a beam. From the literature, the

method of modal superposition is still popular and authors such as Chatterjee et al.

(1994), Cheng et al. (1999), Greco & Santini (2002) and Deng & Cai (2009) often use

it to solve simple beam problems. In Yang & Lin (2005), the method of modal

superposition is used to derive the solution of a vehicle-bridge system that can

simulate a sprung mass traversing a simply support bridge. In their study, they found

that the first mode of vibration, of the simply supported bridge, gave sufficiently

accurate results for the mid-span deflection of the bridge. However, if they had

considered examining internal forces, many more modes of vibrations would have

been needed.


31

2.4 Modelling Irregularities

The greatest amount of damage and impact loads incurred by the track structure or

railway vehicle can be related to track and wheel irregularities. Certain irregularities

can even lead to wheels losing contact with the rail and as the wheels regain contact

with the rail, it can cause large dynamic loads to be applied to a structure, while other

effects can cause a railway vehicle to pitch and roll, which can lead to an undesirable

riding comfort felt by passengers.

In this section, one focuses both on the wheel and track irregularities and examines

their dynamic impact on the bridge and vehicle from the literature.

2.4.1 Wheel Irregularities

According to Esveld (2001), the largest dynamic load experienced by the track

structure from the vehicle results from irregularities on the wheel, such as wheel flats.

The wheel flats are the consequence of wheels sliding as a vehicle applies it brakes. In

the literature, researchers such as Fryba (1996) represent the wheel flat as a periodic

indentation on the rail surface; whereby, the irregularity function is computed using

the depth and length of the wheel flat, the wheel circumference and the distance of the

first impact from the point of origin. Zhai et al. (2001), Sun & Dhanasekar (2002) and

Wu & Thompson (2004) adopt a similar procedure.

2.4.2 Rail Irregularities

As a wheel rolls along the track, it can experience both vertical and lateral

imperfections, which can cause the wheels to vibrate and a vehicle to pitch or roll.

These imperfections can be the result of wear, rail joints, bumps, subsidence and

insufficient maintenance.


32

Perhaps the most commonly used procedure for simulating random rail irregularities

amongst researchers, institutes and organizations is the power spectral density PSD

function. This procedure produces random rail irregularities based on the condition of

the track structure, which is described in great detail in Garg & Dukkipati (1984),

Fryba (1996) and ISO-8606 (1995). Using the PSD function, rail irregularities can be

generated by means of the inverse Fourier transform as shown by Song et al. (2003).

Alternatively, one can simulate rail irregularities, such as rail corrugations and track

misalignment by means of periodic deterministic sinusoidal functions. In Xia et al.

(2000), the lateral irregularity of the track is described by means of a sine function

with an additional random phase angle. The summation of several sine functions is

used to describe the profile of the rail surface in Chang & Lin (1996), while Yau et al.

(1999) adds an extra exponential term that allows the rail irregularities to be applied

gradually to the rail surface.

It is an observation amongst many researchers, such as Yau et al. (1999) and Zhang et

al. (2001) that there is often little difference in the dynamic response of the bridge

regardless of the presence of irregularities; whereas the dynamic response of the

vehicle can be quiet substantial. Wu & Yang (2003) remark that the presence of track

irregularities can have a significant affected on the riding comfort of the train.

2.5 Wheel-rail contacts

This section examines the relationship that exists between the wheel and the rail i.e.

the contact. In finite element programs, such as ANSYS, contact elements are used to

simulate the wheel-rail interaction. Alternatively, many researchers have developed

their own contact systems using the fundamental principles of mechanics.


33

2.5.1 ANSYS contact elements

Within ANSYS, there are several different types of contact elements available, such

as node-to-node, surface-to-surface and node-to-surface contact elements; however,

the latter type is used extensively throughout this thesis to validate different

developed systems.

The node-to-surface contact element defines two regions that may come into contact

with each other: a rigid contact node (wheel) and a flexible target surface (bridge or

rail). The target surface is defined as a straight line between two sets of nodes.

During the solution phase, the contact node penetrates the target surface, proportional

to the contact stiffness; hence, the contact force that exists between the two surfaces is

equal to the penetration times the contact stiffness (ANSYS Theory Reference, 2002).

Contact can only occur when the contact node penetrates the target surface. In order to

determine if a contact node is near the target surface, a pinball region is located

around each target surface. The pinball region is defined by a circle, defaulted at 1.5

times the target surface length. Pseudo elements are then used if a contact node is

located between several target surfaces. These elements determine which target

surface is in contact with the contact node.

From an examination of the node-to-surface contact elements in ANSYS

(CONTAC48 and CONTAC49), it was discovered that these elements had some

limitations, such as:

Contact elements are unable to simulate a wheel or rail with irregularities.

Three-dimensional contact elements required a surface area instead of a line.


34

One is unable to model a contact node (wheel) on an infinite long rigid rail.

Friction models are available but are unsuitable for modelling the accelerating

and decelerating effects of the contact nodes (wheels).

2.5.2 Hertz contact theory

According to Esveld (2001), Hertz contact theory is based on the principal that the

elastic deformation of a steel wheel and a rail create an elliptical contact area. From

the normal force acting on the contact area, one can determine the dimensions of the

contact ellipse, whereby the ratio of the ellipse semi-axes are dependent on the wheel

curvature and rail profile. Typical values of wheel curvature, rail profile and the semi-

axes subjected to a wheel load are presented in Esveld (2001).

As described in Section 2.2.2.3, the Hertzian contact stiffness kH becomes an

important constant in railway dynamics when one assumes that bridge deflection

directly in contact with the wheel is not equal to the deflection of the wheel. Esveld

(2001) computes the Hertzian contact stiffness as a combination of the wheel load,

rail stiffness, radius of wheel curvature, railhead radius and the Poisson ratio.

Alternatively, the Hertzian contact stiffness can be described by a constant depending

on the radii and material properties multiplied by the wheel load. Using Esveld (2001)

one computes that a 1 m wheel diameter and a wheel load of 75kN has a Hertzian

contact stiffness kH = 1.4 GN/m.

2.5.3 Vehicle-bridge interaction element

Throughout the literature, there are many authors such as Yau et al. (1999), Yang &

Wu (2001), Ju & Lin (2003), Song et al. (2003), Wu & Thompson (2004) and Liu et

al. (2008) that have developed their own vehicle-bridge interaction elements or


35

contact systems in the study of railway dynamics. The greatest difference between

each system generally lies in the development of wheel-rail relationship. Some

authors adopt the notion that the wheel of the vehicle never parts with the bridge, thus

including both the local and convective acceleration of the wheel, while other authors

simulate the wheel-rail relationship using a Hertzian contact spring. With the latter

system, the wheel can lose contact with the rail when the contact force becomes

tensile.

The vehicle-bridge interaction element is a means of tracking the wheel-rail

relationship. At each new time step, the vehicle advances a known distance along the

bridge, which is dependant on the speed of the vehicle. As the vehicle advances, the

bridge deflects due to the vehicle load. The bridge and vehicle/wheel displacements

are then used to compute a new contact force. This contact force then imparts a force

to both the vehicle and bridge before proceeding to the next time-step. The force

applied to the vehicle can generate an excitation of the suspension system and causes

the sprung masses of the vehicle to undergo the vertical, longitudinal and lateral

motions depending on their mass, stiffness and damping properties (Majka, 2006).

2.6 Summary

In this chapter the author has introduced and described different aspects to railway

dynamics related to this research project. One initially begins by examining the

vehicle types and their characteristics, which leads onto a brief discussion of the

bridge models that one is most considered about studying and their characteristics.

Next the author describes the effects of irregularities and finally the wheel-rail contact

interaction is examined. Many ideas and concepts discussed in this literature review

related to the author’s study are expanded in greater detail in the following chapters.


36

Chapter 3 - Wheel forces represented as time varying nodal forces and moments

37

Chapter 3

Wheel forces represented as time varying nodal

forces and moments

3.1 Introduction

In this chapter, the author models the wheels of a vehicle as a series of moving forces

traversing a railway bridge; where each wheel is represented as a moving constant

force. As described in the literature, the earliest vehicular models comprise a train

modelled as moving constant forces i.e. only the weights of the vehicle were taken

into account. Fryba (1996: p. 48) concludes that the inertia effects of the vehicles can

be ignored if the mass of the bridge is substantially greater than the mass of the

vehicle, e.g. large spanning bridges.

Modelling the train as a series of moving forces offers many advantages over the

moving sprung and unsprung mass systems, especially if one is only concerned with

the dynamic behaviour of the bridge structure. Since the train is modelled as a set of

forces, there is no external bridge-vehicle interaction required; thus the execution time

of the solution is often quicker. In addition, the results obtained from the moving

force system are often comparable with the results from the moving sprung and


38

unsprung mass systems, even at high speeds (Yang et al., 1997: p. 941; Yang & Wu,

2001: p. 460); thus, modelling a train as a series of moving forces is a recommendable

system of analysing a bridge in the initial stages of any project.

During the early stages of this project, it was discovered that the ANSYS finite

element program would prove to be problematic in the simulation of a moving point

force crossing a beam, as the user was unable to apply a point force between

consecutive nodes; thus, an alternative system was conceived. This alternative system

involves applying nodal forces (approximate simple model) or nodal forces and nodal

moments (exact numerical model) as function of time to the two nodes belonging to

the element on which the moving point force was located.

In order to validate these developed systems within the ANSYS finite element

program, the author introduces the ANSYS node-to-surface contact element

(CONTAC48), where the contact node represents the wheel and the target surface

represents the beam. These contact elements track the horizontal and vertical position

of the contact node in relation with the target surface. To represent the wheel as a

moving force, a point mass element (MASS21) i.e. the wheel, is given a zero mass

and a point force is attached to the node of the point mass element. This node is also

the contact node. In addition, the author must also prescribe a horizontal displacement

to the contact node to simulate the moving force traversing the beam.

Next, the single moving force for the simple model is expanded to simulate several

moving forces traversing a bridge. Originally, to ensure that a node was affected by

only one point force at a time, the distance between wheels of a vehicle had to be


39

greater than twice the length of a finite element; however, the drawback to this

technique is that a large number of nodes are required to discretize the beam.

Therefore an alternative system whereby a node could be affected by several point

forces at any instant was later developed. The overlapping system involves the

summation of several forces as a function of time on a particular node; hence, the

nodal force as a time function is no longer simple but quite complex.

Following the validation of each model, the author examines both a single as well as

several moving forces traversing the Boyne Viaduct railway bridge at a wide range of

speeds. Additionally, the results are made dimensionless by dividing the dynamic

solution by its static loaded solution; this ratio is called the dynamic coefficient. As an

alternative to the single track structure, the author also presents a fictional twin-track

structure based on the aesthetic shape of the Boyne Viaduct to investigate the dynamic

effects of a pair of trains traversing the structure at a range of different speeds. As a

final section, the Boyne Viaduct is subjected to rail traffic prescribed by the

Eurocodes (EN 1991-2, 2003).


40

3.2 Single Moving Force as a Function of Time

3.2.1 Development of single moving force

Equation (A.22), as derived in Appendix A.4, is the differential equation of motion of

an elastic beam subjected to a moving force with EI constant (subscripts dropped) as:

( )4 2

4 2

( , ) ( , ),

v x t v x tEI m p x t

x t

∂ ∂+ =

∂ ∂ (3.1a)

whereby

( ) ( ),p x t x ct Pδ= − where 0L

tc

≤ ≤ (3.1b)

where x is the distance coordinate with the origin at the left-hand end of the beam; t is

the time coordinate with the origin at the instant the force arrives on the beam; v(x, t)

is the deflection of the beam at point x and time t; E is the Young’s modulus of the

beam; I is the constant moment of inertia of the beam cross section; m is the constant

mass per unit length of the beam; P is the concentrated force of constant magnitude, c

is the constant speed of the load motion and L is the total length of the beam as shown

in Figure 3.1. The load is travelling from left to right.

L

P

ct

Figure: 3.1: Simply supported beam subjected to a single moving force

x

y, v

0


41

It should be noted that the first term on the left hand side of Equation (3.1a) describes

the beam stiffness while the second term is the beam inertia, whereas the term on the

right hand side of Equation (3.1a) describes the moving load.

The symbol δ(x- x0) is the Dirac (impulse, or delta) function and can be defined as the

distributional derivative of the Heaviside function ( )0x x−H as presented in Figure

3.2 (a) and (b), respectively. The Dirac delta function is defined as:

( ) ( ) ( )0 0

0

L

f x x x dx f xδ − =∫ so long as 00 x L< < (3.2)

Figure 3.2: Plot of the mathematical functions: (a) Dirac delta function; (b)

Heaviside Function

Recalling Equation (B.10) in Appendix B, the author multiples Equation (3.1a) by a

weighting function vδ and integrates along the local length of a finite element, where

the local coordinate is defined by the symbol χ , giving:

( )4 2

4 2

0

, 0

lv v

EI m p x t vdt

δ χχ

∂ ∂+ − =

∂ ∂ ∫ (3.3)

(a) (b)

Note:

0ε →

0x

( )xδ

ε

0x0

2x

ε+

x x

00

11ε

02

xε

−

( )xH


42

Integrating Equation (3.3) by parts twice, as shown in Equation (B.13) gives:

2 2 2

2 2 2

0

lv v v

EI m v dt

δδ χ

χ χ

∂ ∂ ∂+

∂ ∂ ∂ ∫

( ) 1 1 2 20

0 0

ˆ ˆˆ ˆ,

l

l

l

v vp x t vd v Q M v Q M

δ δδ χ δ δ

χ χ

∂ ∂= + + + +

∂ ∂∫ (3.4)

where 1

Q and 2

Q are the nodal shear forces while 1

M and 2

M are the nodal bending

moments at local nodes 1 and 2 of a beam element, respectively.

Next, the weighting function vδ is made equal to a beam shape function. The result is

Equation (B.15) in Appendix B. This equation is rewritten below with the right-hand

side fully evaluated as shown by Equation (B.18) as:

( )( )( )( )

( )( )( )( )

1 1

2 21 1

2 2

2 20 0

2 2

l l

N N

G Gv vEI d m d

N N t

G G

χ χ

χ χχ χ

χ χχ

χ χ

′′ ′′ ∂ ∂

+ ′′ ∂ ∂

′′

∫ ∫

( )( )( )( )

( )

11

1 1

20 2

22

ˆ

ˆ,

ˆ

ˆ

l

QN

G Mp x t d

N Q

G M

χ

χχ

χ

χ

= +

∫ (3.5)

where ( )1N χ , ( )1G χ , ( )2N χ and ( )2G χ are the weighting functions, which are

equal to the shape functions that are defined by Equation (B.8) and plotted in Figure

B.3 in Appendix B. Equation (3.5) will be known as the exact numerical model.

However, one begins by simplifying the first term on the right hand side of Equation

(3.5). This is done by replacing ( )1N χ and ( )2N χ by ( )1H χ and ( )2H χ ,

respectively, where ( )1H χ and ( )2H χ are linear weighting functions. Furthermore,

( )1G χ and ( )2G χ are replaced by zeros. With these changes Equation (3.5) becomes:


43

( )( )( )( )

( )( )( )( )

( )

( )( )

11 1 1

2 21 1 1

2 2

2 2 20 0 0 2

2 22

ˆ

ˆ0,

ˆ

0 ˆ

l l l

QN N H

G G Mv vEI d m d p x t d

N N t H Q

G G M

χ χ χ

χ χχ χ χ

χ χχ χ

χ χ

′′ ′′ ∂ ∂

+ = + ′′ ∂ ∂

′′

∫ ∫ ∫ (3.6)

where ( )1H χ and ( )2H χ are defined in Equation (B.47) and can be seen in Figure

B.7 in Appendix B. The solution to Equation (3.6) will be known as the simple model.

In Figure 3.3, the coordinate of node i is xi, which is the distance of node i from the

left hand support. Time t is arranged in such a manner that the point force is over the

left hand support at t = 0.

+1-1 iii

ct

x x x x

3

x

2

x

1

x =

P

L

2 3 i i+i- 11 L1 0

Figure: 3.3: Simply supported beam subjected to a single moving force

The purpose of Sections 3.2.1.1 and 3.2.1.2 is to express the effect of the applied force

P at node i, by means of the simple and exact model, respectively. To affect node i,

the force P must be applied to one of two finite elements, one element to the left of

node i, for which node i is its local node 2 and the other element to the right of node i,

for which node i is its local node 1.

x

y, v


44

3.2.1.1 Moving force using an approximate simple model

One begins by examining the finite element to the left of node i. Figure 3.4 presents

an isolated beam element from node i-1 to node i with the point force P between those

two nodes. The distance between nodes xi-1 and xi is l, while the distance from xi-1 to a

typical point of a beam element is defined as χ . The internal shear force and bending

moment at the ends of the beam are defined as Q and M , respectively.

Figure 3.4 An isolated finite element from node i-1 to node i (simple)

The global x position of a typical point on the isolated beam element in Figure 3.4 can

be related to the local χ position as follows:

1ix xχ −= + (3.7)

Substituting Equation (3.7) into (3.1b) gives the local x position of the moving force P

on this isolated beam element as:

( ) ( ) ( ) ( )( )1 1, i ip x t x ct P x ct P ct x Pδ δ χ δ χ− −= − = + − = − − (3.8)

which is only valid for 1i ix x

tc c

− ≤ ≤

( )1 1iPH ct x −− ( )2 1iPH ct x −−

1Q2Q

1M 2M

χ


45

Substituting Equation (3.8) into the first term on the right hand side of Equation (3.6)

relates the pressure term ( ),p x t to the moving force P as follows:

( )

( )( )

( )

( )( )( )

1 1

1

2 20 0

0 0,

0 0

l l

i

H H

p x t d ct x PdH H

χ χ

χ δ χ χχ χ

−

= − −

∫ ∫

( )

( )

1 1

2 1

0

0

i

i

H ct x

PH ct x

−

−

−

= −

(3.9)

where the right hand side of Equation (3.9) is evaluated using Equation (3.2).

Recalling Equation (B.47) in Appendix B, the linear shape functions (positive

between 0 lχ< < ) are defined as:

( )1 1H lχ χ= −

( )2H lχ χ= (3.10)

Substituting Equation (3.10) into (3.9) gives

( )

( )

1

1 1

12 1

1

00

0

0

i

i

ii

ct x

H ct x l

P Pct xH ct x

l

−

−

−−

− − −

= −−

for 1i ix x

tc c

− ≤ ≤ (3.11)

Equation (3.11) states that when the point force P is between the nodes i-1 and i, it

can be replaced by two nodal forces acting in the same direction as the force P, one

applied at node i-1 and the other at node i, where both are a function of time.


46

Now, one examines the element to the right of node i. An isolated beam element from

node i to node i+1 is selected, when the point force P is between these two nodes as

shown in Figure 3.5. On this occasion, the global x position of a typical point on the

isolated beam element in Figure 3.5 can be related to the local χ position as follows:

ix xχ= + (3.12)

Figure 3.5: An isolated finite element from node i to node i+1 (simple)

Substituting Equation (3.12) into (3.1b) gives the local x position of the moving force

P on this isolate beam element as:

( ) ( ) ( ) ( )( ), i ip x t x ct P x ct P ct x Pδ δ χ δ χ= − = + − = − − (3.13)

which is valid for 1i ix x

tc c

+≤ ≤ . Substituting Equation (3.13) into the first term on the

right hand side of Equation (3.6) and evaluating using Equation (3.2) gives:

( )

( )( )

( )

( )( )( )

1 1

2 20 0

0 0,

0 0

l l

i

H H

p x t d ct x PdH H

χ χ

χ δ χ χχ χ

= − −

∫ ∫

( )

( )

1

2

0

0

i

i

H ct x

PH ct x

−

= −

(3.14)

( )1 iPH ct x− ( )2 iPH ct x−

1Q 2Q1M 2M

χ


47


( )

( )

1

2

1

00

0

0

i

i

ii

ct x

H ct x l

P Pct xH ct x

l

− − −

= −−

for 1i ix x

tc c

+≤ ≤ (3.15)

where Equation (3.15) tells the reader that when the point force P is between the

nodes i and i+1, it can again be replaced by two nodal forces acting in the same

direction as the force P, one applied at node i and the other at node i+1, where both

are a function of time.

The function between 1ix

c

− and ix

c in Figure 3.6 is the third entry of the vector on the

right hand side of Equation (3.11), while the function between ix

c and 1i

x

c

+ in Figure

3.6 is the first entry of the vector on the right hand side of Equation (3.15)

Figure 3.6: Influence of point force P on node i as a function of time

t

P

1ix

c

+ix

c

1ix

c

−0

1 ict xP

l

− −

1ict xP

l

−−


48

3.2.1.2 Moving force using an exact numerical model

Using a similar methodology to that of Section 3.2.1.1, the author begins by

examining the element to the left of node i. An isolated beam element from node i-1

to node i is selected when the point force P is between these nodes as shown in

Figure 3.7.

Figure 3.7 An isolated beam element from node i-1 to node i (exact)



( )( )( )( )

( )

( )( )( )( )

( )( )

1 1

1 1

1

2 20 0

2 2

,

l l

i

N N

G Gp x t d ct x Pd

N N

G G

χ χ

χ χχ δ χ χ

χ χ

χ χ

−

= − −

∫ ∫

( )( )( )( )

1 1

1 1

2 1

2 1

i

i

i

i

N ct x

G ct xP

N ct x

G ct x

−

−

−

−

−

− =

− −

(3.16)

where the right hand side of Equation (3.16) is evaluated using Equation (3.2).

Recalling Equation (B.8) in Appendix B, the shape functions (between 0 lχ< < ) are

defined as:

( )2 1iPG ct x −−( )1 1iPG ct x −−

( )2 1iPN ct x −−( )1 1iPN ct x −−

1Q 2Q1M2M

χ


49

( )2 3

1 2 3

3 21N

l l

χ χχ = − +

( )2 3

1 2

2G

l l

χ χχ χ= − +

( )2 3

2 2 3

3 2N

l l

χ χχ = −

( )2 3

2 2G

l l

χ χχ = − + (3.17)


( )( )( )( )

( ) ( )

( )( ) ( )

( ) ( )

( ) ( )

2 3

1 1

2 3

2 31 1 1 1

1 21 1

2 32 1

1 1

2 32 1

2 3

1 1

2

3 2 1

2

3 2

i i

i i i

ii

ii i

i

i i

ct x ct x

l l

N ct x ct x ct xct x

G ct x l lP P

N ct x ct x ct x

G ct x l l

ct x ct x

l l

− −

− − −

−−

−− −

−

− −

− −− +

− − −− − +

− =

− − − − −

− −− +

for 1i ix x

tc c

− ≤ ≤ (3.18)

where Equation (3.18) states that when the point force P is between the nodes i-1 and

i, it can be replaced by two nodal forces as well as two nodal moments, one force and

one moment applied at node i-1 and the other force and moment at node i, where all

are functions of time. It should be noted that the nodal forces applied act in the same

direction as the positive y-axis when positive, while the nodal moments applied act

counter-clockwise about the z-axis when positive, as shown in Figure 3.7.

Next an element to the right of node i is examined. In Figure 3.8 an isolated beam

element from node i to node i+1 is selected when the point force P is between these

particular nodes.


50

Figure 3.8: An isolated section of the beam from node i to node i+1 (exact)



( )( )( )( )

( )

( )( )( )( )

( )( )

1 1

1 1

2 20 0

2 2

,

l l

i

N N

G Gp x t d ct x Pd

N N

G G

χ χ

χ χχ δ χ χ

χ χ

χ χ

= − −

∫ ∫

( )( )( )( )

1

1

2

2

i

i

i

i

N ct x

G ct xP

N ct x

G ct x

−

− =

− −

(3.19)

While substituting Equation (3.17) into (3.19) gives:

( )( )( )( )

( ) ( )

( )( ) ( )

( ) ( )

( ) ( )

2 3

2 3

2 31

21

2 32

2 32

2 3

2

3 2 1

2

3 2

i i

i i i

ii

ii i

i

i i

ct x ct x

l l

N ct x ct x ct xct x

G ct x l lP P

N ct x ct x ct x

G ct x l l

ct x ct x

l l

− − − + − − − − − +

− =

− − − − −

− −− +

for 1i ix x

tc c

+≤ ≤ (3.20)

As before, Equation (3.20) states that when the point force P is between nodes i and

i+1, it can again be replaced by two nodal forces as well as two nodal moments, one

force and moment applied at node i and the other force and moment at node i+1,

where all are functions of time.

( )2 iPG ct x−( )1 iPG ct x−

( )2 iPN ct x−( )1 iPN ct x−

1Q 2Q1M 2M

χ


51

The functions between 1ix

c

− and ix

c in Figure 3.9 (a) and (b) are the third and fourth

entries of the vector on the right hand side of Equation (3.18), respectively, while the

functions between ix

c and 1i

x

c

+ in Figure 3.9 (a) and (b) are the first and second entries

of the vector on the right hand side of Equation (3.20), respectively.

Figure 3.9: Influence of point force P on node i (a) nodal force (b) nodal moment

(a)

(b)

1ix

c

+

1ix

c

+

ix

c

ix

c

1ix

c

−

1ix

c

−

t

t0

0

P

( ) ( )2 3

1 1

2 3

3 2i i

ct x ct xP

l l

− − − −

−

( ) ( )2 3

2 3

3 2 1

i ict x ct x

Pl l

− −− +

( )( ) ( )

2 3

2

2i i

i

ct x ct xP ct x

l l

− −− − +

( ) ( )2 3

1 1

2

i ict x ct x

Pl l

− − − −

− +


52

3.2.1.3 Moving single force using ANSYS node-surface contact elements

In Bowe & Mullarkey (2000: p. 46), a wheel of a vehicle is represented by a single

point mass element Mw located at its own unique node im as can be seen in Figure

3.10. Node-to-surface contact elements (CONTAC48) are then used to simulate the

wheel-rail interaction: whereby, node im is referred to as the contact node and the

nodes between x1 to xL of the beam are referred to as the target surface. The contact

element 2 and i are also shown in Figure 3.10.

+1

i+1x

i-1

contact

Lx

i

i

1 ix

contact node

i

L

x

2 3

32x x

i m

target surface

1

1x = 0

Mwelement 2

i-

contact element

Figure 3.10: Single point mass element with NL-1 contact elements

In order to represent the wheel as a moving load, the point mass element Mw is given

zero mass and a point force P is attached to im, the node of the point mass, as

illustrated in Figure 3.11. Within ANSYS, any node can be given a prescribed

displacement that varies with time. Therefore, the point mass is moved across the

beam from left to right by prescribing the longitudinal displacement, U = ct, to node

im, where c is the speed of the point mass element. This system is primarily used to

validate the simple and exact numerical model.

U = ct

Mw

P

Figure 3.11: Massless wheel with a point force P attached

x

x

y, v


53

3.2.2 Validation of moving force as a function time

In order to validate simple and exact numerical system, the author compares results

from these techniques with the results of the ANSYS contact element method as well

as with results from the literature.

3.2.2.1 Cantilever beam subjected to a moving load

In this first example, one simulates a moving load traversing a cantilever beam from

the fixed to the free-end, and vice-versa, where the beam has a length L of 7.62 m,

flexural rigidity EI of 9.47 x 106 Nm

2 and mass per unit length m of 46 kg/m;

therefore, using Equation (C.27a), the first natural frequency of this particular beam is

1 27.49 rad/sec.ω = The moving load P of -25.79 kN traverses the beam at a constant

speed c of 50.8 m/s. These properties are the same as those used by Akin & Mofid

(1989). The gravitational and damping effects of the beam are ignored. In all cases,

time t is arranged in such a manner that the load is at the left support at t = 0 sec and

the initial displacement and velocity of the beam are equal to zero. The Newmark-β

time integration scheme (Bathe, 1996) with 100 equal time steps is used to solve the

transient analysis. The beam is discretized into ten beam elements.

The vertical displacement at the free-end of the fixed-free and free-fixed cantilever

beam subjected to a moving load, traversing from left to right, can be seen in Figure

3.12a and 3.12b, respectively. It can be observed from the results that both the

solutions of both the simple system and the exact system are comparable with the

numerical results obtained from Akin & Mofid (1989).


54

-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

Simple Model

Exact Model

Akin & Mofid (1989) ML

-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

Simple Model

Exact Model

Akin & Mofid (1989) ML

Figure 3.12: Time history at free-end of a cantilever beam: (a) left hand side is

fixed; (b) right hand side is fixed.

3.2.2.2 Simply supported beam subjected to a moving load

In this second example, the beam and vehicle properties adopted are similar to those

of Yang and Wu (2001), such that the bridge has a length L of 25m, Young’s modulus

of elasticity E of 2.87x106 kN/m

2, moment of inertia I of 2.9 m

4, mass per unit length

m of 2.303 t/m and a Poisson’s ratio ν of 0.2. The gravitational and damping effects of

the bridge are ignored. The vehicle traverses the bridge at a constant speed c of 27.78

m/s (100 km/hr) and has a vehicle load P of -56.4 kN. Again one uses the Newmark-β

(a)

(b)


55

time integration method (Bathe, 1996) with 500 equal time steps to solve each

transient analysis, with the vehicle at the left hand support at t = 0 sec. In addition, the

initial displacement and velocity of the beam are equal to zero at time t = 0 sec. As

before, the beam is discretized into ten elements.

In Figure 3.13(a) the vertical displacement of the beam at mid-span is presented. It

can be seen that there is a great likeness among the solutions of all three systems as

the graphs have identical magnitude and curvature. Examining the vertical

acceleration, ( )2 22 ,v L x∂ ∂ at mid-span of the beam, it is seen that the results in all

cases are also quite similar as shown in Figure 3.13(b). It is observed that the vertical

acceleration of the beam range between -0.3 to 0.4 m/s2 in all cases. In the final plot, the

bending moment at mid-span of the beam as a function of time is presented in Figure

3.13(c). The reader should be aware that in the case of the exact model, the bending

moment at mid-span of the beam varies between the two local nodes at mid-span. This

is due to the applied nodal moment. Hence, results from both local nodes are plotted in Figure

3.13c for the exact model. Similarly to the previous graphs, the simple model gives results which

compare well with results from both the exact model and the use of the ANSYS contact element.

-0.0025

-0.0020

-0.0015

-0.0010

-0.0005

0.0000

0.0005

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

Simple Model

Exact Model

ANSYS CE

(a)


56

-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

0.4

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al A

cc

ele

rati

on

(m

/s2)

Simple Model Exact Model ANSYS CE

Figure 3.13: (a) Vertical displacement; (b) vertical acceleration; (c) bending

moment at mid-span of the beam as a function of time

3.2.2.3 Sensitivity analysis of the developed systems

As seen in the previous results, both the simple model and the exact model have

identical results provided that a suitable number of beam elements are used to

discretize the beam. However, by selecting a coarser beam mesh, i.e. only 4 beam

elements as presented in Figure 3.14, it will be shown that the results from the simple

model can be different from those of the exact model.

(b)

(c)

Point load

arrives at node 4 Point load

arrives at node 6

Point load

arrives at node 5


57

L = 25m

1 2 3 4 5Element 1 Element 2 Element 3 Element 4

LN2 LN1 LN2 LN1 LN2 LN1 LN2LN1

l = 6.25m

Figure 3.14: Simply supported beam from Section 3.2.2.2 divided into four elements

In Figure 3.15, the author plots the vertical displacement, acceleration and bending

moment at mid-span of the beam as a function of time for all three methods. For

comparison purposes, the author also presents results from the ANSYS contact model,

where the beam is discretized into 50 elements.

The first noticeable distinction observed in Figure 3.15, is that the results obtained

from the ANSYS node-to-surface contact element, with 4 elements, (CONTAC48) are

almost identical to the results from the simple model, while the second distinction is

that these particular results are somewhat lower than the results from the exact

numerical model and the ANSYS contact model with 50 elements. Additionally, from

inspection of Figure 3.15, one can conclude that the results from the exact model tend

to give better comparisons with the more accurate solutions in Figure 3.13.

It can be seen in Figure 3.15(a) that the maximum mid-span deflection of the simple

model is at least 10% smaller than that of the exact model, while the vertical

acceleration for simple model tends to diverge with increasing time in Figure 3.15(b).

In addition, the bending moment for the simple model in Figure 3.15(c) tends to be

slightly less than the bending moment for the simple model observed in Figure

3.13(c). The reader can see that despite the exact model of Figure 3.15(a) having an

identical deflection with the exact model in Figure 3.13(a), its bending moments in

Figure 3.15(c) are quite different from the exact model’s bending moment of Figure

3.13(c), using the coarser beam mesh.


58

-0.0025

-0.0020

-0.0015

-0.0010

-0.0005

0.0000

0.0005

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al D

isp

lac

me

nt

(m)

Simple Model (4 elements)

Exact Model (4 elements)

ANSYS CE (4 elements)

ANSYS CE (50 elements)

-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

0.4

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al A

cc

ele

rati

on

(m

/s2)

Simple Model (4 elements) Exact Model (4 elements)

ANSYS CE (4 elements) ANSYS CE (50 elements)

0

100

200

300

400

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Be

nd

ing

Mo

me

nt

(kN

m)

Simple Model

Exact Model (LN1)

Exact Model (LN2)

ANSYE CE (4 elements)

ANSYE CE (50 elements)

Figure 3.15: Time history at mid-span of the beam using only 4 beam elements: (a)

vertical displacement; (b) vertical acceleration; (c) bending moment

(a)

(b)

(c)

Point load

arrives at node 2 Point load

arrives at node 4

Point load

arrives at node 3


59

In Figure 3.16, the author presents two free-body diagrams at the mid-span joint (i=3)

of the beam. One is when the moving point force is on the second element, while the

other is when the moving point force is on the third element. By satisfying joint

equilibrium in Figures 3.16(a) and (b) and using Figure B.2, the internal bending

moments at mid-span of the beam are equal to ( )2 1ˆ ˆ 1,2 .iM M PG i− = = Thus, the

difference or gap between the two internal bending moments is equal to the applied

moment at that joint for the exact model as shown in Figure 3.17. Equally, the internal

shear forces at mid-span of the beam are equal to ( )2 1ˆ ˆ 1, 2 .iQ Q PN i− = =

(a) Internal forces in node 3 when the moving point force is on the 2nd

element

(b) Internal forces in node 3 when the moving point force is on the 3rd

element

Figure 3.16: Free-body diagram of the joint at mid-span of the beam

P

2PG

Node 3

2M1M

2Q1Q

2M1M

1Q 2Q

2PN1PN

1PG

2i = 3i =Elem 2 Elem 33i = 4i =

2M1M

2Q1Q

2M1M

1Q 2Q

2PN1PN

1PG

2i = 3i =Elem 2 Elem 33i = 4i =

P

2PG

Node 3

2Q 1Q

1M2M

3i =

2Q 1Q

1M2M

3i =


60

-100

0

100

200

300

400

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Be

nd

ing

Mo

me

nt

(kN

m)

Internal Moment (LN2)

Internal Moment (LN1)

Applied Moment

Figure 3.17: Time history of the internal and applied moments at mid-span of the

beam using the exact model with only 4 beam elements

3.2.2.4 Examining internal forces of the developed systems

Like all finite element programs, ANSYS has it limitations. As mentioned earlier in

this chapter, ANSYS does not allow the user to apply a point force between

consecutive nodes; however, using the author’s developed system this problem can be

overcome. Nonetheless, these developed systems have certain limitations within

ANSYS, which arise when one examines the internal forces. In the following

example, the author adopts the same beam properties as those of Section 3.2.2.3, such

that the beam has a total length L of 25 m; therefore, the distance between consecutive

nodes of a single element l is 6.25 m. In Figure 3.18, a point force P is applied

between two consecutive nodes at a distance 8.75 mx = from the left support or a

distance 0.4l from the second node.

Load P on Elem 1 Load P on Elem 2 Load P on Elem 3 Load P on Elem 4

2 PG

1 PG


61

x

y

l

P

1 3 4 52

0.4 0.6ll

x = 8.75m

L = 25m

Figure 3.18: Applying a point force between two consecutive nodes in ANSYS

Using ( )0.4 0.4 6.25 2.5 ml = = with Equation (B.8) and (B.47), the author is able to

compute the shape functions that are used by the simple and exact models as follows:

( )1 2.5 0.600H = ( )2 2.5 0.400H =

( )1 2.5 0.648N = ( )2 2.5 0.352N =

( )1 2.5 0.900G = ( )2 2.5 0.60G = − (3.21)

Using the computed values in Equation (3.21), the author is now able to replace the

original point force in Figure 3.18 with two nodal forces 0.6P and 0.4P applied at

nodes 2 and 3, respectively, for the simple model as shown in Figure 3.19(a). In a

similar manner, the original point force in Figure 3.18 can be replaced by two nodal

forces and two nodal moments for the exact model as shown in Figure 3.19(b). The

nodal forces applied to nodes 2 and 3 are 0.648P and 0.352P, respectively, while the

nodal moments are 0.9P and -0.6P as indicated on the diagram. As an additional

exercise, the author also considers replacing the original point force in Figure 3.18

with a pressure applied to the beam such that the centre position of the pressure is

Element 1 Element 3 Element 4

Element 2


62

located at a distance of 0.4l from the second node and the pressure has an overall

length of 0.1l, as can be seen in Figure 3.19(c).

0.6P 0.4P

-0.65P -0.35P

0.648P 0.352P

-0.65 P -0.35P

- 0.6 P0.9 P

-0.65P -0.35P

lp=P0.1

Figure 3.19: Point force represented in ANSYS: (a) simple model; (b) exact model;

(c) pressure load

Using Figure 3.19(a), the resultant reaction forces experienced by the beam are as

follows: the left-hand support or LHS ( ) ( )0.6 3 4 0.4 2 4 0.65P L P L L P= − + = − ; thus,

giving the RHS = –0.35P as illustrated in Figure 3.19. It should also be noted that if

one had chosen Figures 3.19(b) or 3.19(c), the resultant reaction forces would be same

as the computed values. In this example, the point force P is 56.4 kN; therefore, with

the aid of Figure 3.19 and Equation (3.21) the nodal forces and moments used in this

example can be computed as follows:

(a)

(b)

(c)


63

( )LHS 0.650 56.4 36.66= − = − ( )RHS 0.350 56.4 19.74= − = −

( ) ( )1 2.5 0.600 56.4 33.84PH = = ( ) ( )2 2.5 0.400 56.4 22.56PH = =

( ) ( )1 2.5 0.648 56.4 36.55PN = = ( ) ( )2 2.5 0.352 56.4 19.85PN = =

( ) ( )1 2.5 0.900 56.5 50.76PG = = ( ) ( )2 2.5 0.60 56.4 33.84PG = − = − (3.22)

The reader should take the time to refer back to Equation (3.22) and Figure A.3 as

they examine the internal shear forces in Figure 3.20. In Figure 3.20a, the author

presents the analytical shear force Qy, when the point force is located between nodes 2

and 3, while Figures 3.20b to 3.20d plot the numerical shear forces that one gets from

a finite element program. Figure 3.20b, 3.20c and 3.20d show the internal shear forces

Qy from the simple model, exact model and pressure model, respectively. From

inspection of Figure 3.20, one can see that the numerical solutions are somewhat

similar to the analytical solution and would improve with a larger number of beam elements.

(a)

(b)

(c)

yQ

x

yQ

yQ

x

x


64

Figure 3.20: Internal shear force diagram: (a) analytical solution; (b) simple

model; (c) exact model; (d) pressure load

The reader should again refer back to Equation (3.22) and Figure A.3 in order to

check the internal bending moments Mz that are presented in Figure 3.21. In Figure

3.21a, the analytical bending moment, when the point force is located between nodes

2 and 3, is plotted. Figures 3.21b to 3.21d then show the numerical bending moment

that one gets from a finite element program. Figure 3.21b, 3.21c and 3.21d are the

bending moments from the simple model, exact model and pressure model, respectively.

(d)

(a)

(c)

(b)

yQ

x

zM

zM

zM

x

x

x


65

Figure 3.21: Internal bending moment diagram: (a) analytical solution; (b) simple

model; (c) exact model; (d) pressure load

Inspecting Figure 3.21, one firstly sees that the bending moment for the simple model

is exactly the same as the pressure model. Secondly, the bending moment between

nodes 2 and 3 for the analytical solution is not captured in any of the numerical

solutions. Thirdly, the exact model has stepped internal moments. These stepped

internal moments are due to the applied moments 1PG and 2PG at nodes 2 and 3. A

free-body diagram of nodes 2 and 3 for the exact model is shown symbolically in

Figure 3.22a and 3.23a, and numerically in Figure 3.22b and 3.23b, respectively. The

numerical values in Figure 3.22b and 3.23b are taken directly from Figures 3.20 and

3.21. However, before applying equilibrium to the joint, one must also refer to Figure

A.3, which shows the sign convention adopted in this thesis. On the negative face, the

shear force is positive downwards and the bending moment is positive in a clockwise

direction, whereas on the positive face, the shear force is positive upwards and the

bending moments are positive in an anti-clockwise direction. Applying equilibrium to

node 2 for the exact model, one gets:

( ) ( ) 1

at element 2 at element 3

0 0y y

Q l Q PN− − =

or 2 1 1ˆ ˆ 0y yQ Q PN− − = ⇒ ( ) ( ) ( )36.66 0.11 36.55 0− − = (3.23a)

( ) ( ) 1


0 0z z

M l M PG− − =

or 2 1 1ˆ ˆ 0

z zM M PG− − = ⇒ ( ) ( ) ( )229.13 279.89 50.76 0− − − − = (3.23b)

(d) zM

x


66

Figure 3.22: Free-body diagram of node 2 (a) symbolically; (b) numerically

Equally, applying equilibrium to node 3 for the exact model, one gets:

( ) ( ) 2


0 0y y

Q l Q PN− − =

or 2 1 2ˆ ˆ 0y yQ Q PN− − = ⇒ ( ) ( ) ( )0.11 19.74 19.85 0− − − = (3.24a)

( ) ( ) 2


0 0z z

M l M PG− − =

or 2 1 2ˆ ˆ 0

z zM M PG− − = ⇒ ( ) ( ) ( )280.59 246.75 33.84 0− − − − − = (3.24b)

Figure 3.23: Free-body diagram of node 3 (a) symbolically; (b) numerically

Despite the stepped internal moments for the exact model, Equations (3.23) and (3.24)

show that this solution is numerically correct as joint equilibrium is satisfied.

36.66 0.11

279.89−

229.13− 36.55

50.76

(a) (b)

0.11 19.74−

246.75−280.59−19.85

-33.84

2ˆ

yQ 1ˆ

yQ

1ˆ

zM

2ˆ

zM

1PN

1 PG

2i = 2i =

(a) (b) 2

ˆyQ

1ˆ

yQ

1ˆ

zM

2ˆ

zM

2PN

2 PG

3i = 3i =


67

Until now, one has computed the internal forces analytically using the beam element

shape functions; however, ANSYS uses the beam stiffness and nodal displacements as

well as the beam shape functions to calculate the internal forces. Recalling Equation

(A.16) and (A.19) in Appendix A, the bending moment and shear force are computed,

respectively, as follows:

2

2z z

vM EI

χ

∂=

∂ (3.25a)

3

3y z

vQ EI

χ

∂= −

∂ (3.25b)

where the second and third derivative of the shape function with respect to x can be

found in Equation (B.8). Table 3.1 presents the bending moment, while the shear

force can be seen in Table 3.2 for the simple model.

Table 3.1: Bending moment using Equation (3.25a) for simple model

Second derivative of shape functions Nodal Displacements

N1'' G1'' N2'' G2'' Elem 1 Elem 2 Elem 3 Elem 4

Local node 1 0.00E+00 1.35E-03 1.79E-03 1.19E-03

-0.1536 -0.6400 0.1536 -0.3200 2.45E-04 1.59E-04 -1.99E-05 -1.59E-04

Local node 2 1.35E-03 1.79E-03 1.19E-03 0.00E+00

0.1536 0.32 -0.1536 0.64 1.59E-04 -1.99E-05 -1.59E-04 -2.05E-04

Bending Moment in element at LN1 (kNm) 0.00 -229.13 -246.75 -123.37

Bending Moment in element at LN2 (kNm) -229.13 -246.75 -123.37 0.00

Table 3.2: Shear force using Equation (3.25b) for simple model

Third derivative of shape functions Nodal Displacements

N1''' G1''' N2''' G2''' Element 1 Element 2 Element 3 Element 4

Local node 1 0.00E+00 1.35E-03 1.79E-03 1.19E-03

0.0492 0.1536 -0.0492 0.1536 2.45E-04 1.59E-04 -1.99E-05 -1.59E-04

Local node 2 1.35E-03 1.79E-03 1.19E-03 0.00E+00

0.0492 0.1536 -0.0492 0.1536 1.59E-04 -1.99E-05 -1.59E-04 -2.05E-04

Shear force in element at LN1 (kN) 36.66 2.82 -19.78 -19.78



68

In a similar manner, Table 3.3 presents the bending moment, while the shear force is

shown in Table 3.4 for the exact model. Comparing the bending moments and shear

forces for the simple model in Tables 3.1 and 3.2 with Figures 3.20b and 3.21b, the

reader can clearly see that both sets of results are identical. Similarly, comparing the

bending moment and shear forces for the exact model in Table 3.3 and 3.4 with

Figures 3.20c and 3.21c, one can again see that both sets of results are alike.

Table 3.3: Bending moment using Equation (3.25a) for exact model

Second derivative of shape functions Nodal Displacements

N1'' G1'' N2'' G2'' Elem 1 Elem 2 Elem 3 Elem 4

Local node 1 0.00E+00 1.48E-03 1.94E-03 1.26E-03

-0.1536 -0.6400 0.1536 -0.3200 2.65E-04 1.79E-04 -3.15E-05 -1.70E-04

Local node 2 1.48E-03 1.94E-03 1.26E-03 0.00E+00

0.1536 0.32 -0.1536 0.64 1.79E-04 -3.15E-05 -1.70E-04 -2.17E-04

Bending Moment in element at LN1 (kNm) 0.00 -279.89 -246.75 -123.37

Bending Moment in element at LN2 (kNm) -229.13 -280.59 -123.37 0.00

Table 3.4: Shear force using Equation (3.25b) for exact model

Third derivative of shape functions Nodal Displacements

N1''' G1''' N2''' G2''' Element 1 Element 2 Element 3 Element 4

Local node 1 0.00E+00 1.48E-03 1.94E-03 1.26E-03

0.0492 0.1536 -0.0492 0.1536 2.65E-04 1.79E-04 -3.15E-05 -1.70E-04

Local node 2 1.48E-03 1.94E-03 1.26E-03 0.00E+00

0.0492 0.1536 -0.0492 0.1536 1.79E-04 -3.15E-05 -1.70E-04 -2.17E-04



Before ending this section, the author will now show that internal forces from

Equation (3.6) and (3.5), for the simple and exact models, are quite similar with the

internal forces for the pressure loaded model shown in Figures 3.20d and 3.21d,

respectively. Recalling Equation (3.6) and (3.5), where the beam inertia has been

omitted as follows:


69

( )( )( )( )

( )

( )( )

11 1

21 1

2

2 20 0 2

22

ˆ

ˆ0,

ˆ

0 ˆ

l l

QN H

G MvEI d p x t d

N H Q

G M

χ χ

χχ χ

χ χ χ

χ

′′ ′′ ∂

− = ′′ ∂

′′

∫ ∫ (3.26a)

( )( )( )( )

1

21

2

20

2

l

N

G vEI d

N

G

χ

χχ

χ χ

χ

′′ ′′ ∂

′′ ∂ ′′

∫

( )( )( )( )

( )

11

1 1

20 2

22

ˆ

ˆ,

ˆ

ˆ

l

QN

G Mp x t d

N Q

G M

χ

χχ

χ

χ

− =

∫ (3.26b)

With the aid of Equations (B.15) to (B.19) in Appendix B, the author is able to rewrite

the 1st term on the left hand side of Equation (3.26) as follows:

1

2 2

1

3

2

2 2

2

12 6 12 6

ˆ6 4 6 2

12 6 12 6

ˆ6 2 6 4

z

z

vl l

l l l lEI

vl ll

l l l l

θ

θ

− −

− − − −

( )

( )( )

11

1

20 2

2

ˆ

ˆ0,

ˆ

0 ˆ

l

QH

Mp x t d

H Q

M

χ

χχ

− =

∫ (3.27a)

1

2 2

1

3

2

2 2

2

12 6 12 6

ˆ6 4 6 2

12 6 12 6

ˆ6 2 6 4

z

z

vl l

l l l lEI

vl ll

l l l l

θ

θ

− −

− − − −

( )( )( )( )

( )

11

1 1

20 2

22

ˆ

ˆ,

ˆ

ˆ

l

QN

G Mp x t d

N Q

G M

χ

χχ

χ

χ

− =

∫ (3.27a)

where the nodal displacements 1 1 2 2ˆ ˆ, , and z zv vθ θ are obtained directly from ANSYS.

Finally, substituting Equations (3.14) and (3.19) into Equation (3.27) gives:

1

2 2

1

3

2

2 2

2

12 6 12 6

ˆ6 4 6 2

12 6 12 6

ˆ6 2 6 4

z

z

vl l

l l l lEI

vl ll

l l l l

θ

θ

− −

− − − −

11

1

2 2

2

ˆ

ˆ0

ˆ

0 ˆ

QH

MP

H Q

M

− =

(3.28a)

[k] [u] [fext] [fint]


70

1

2 2

1

3

2

2 2

2

12 6 12 6

ˆ6 4 6 2

12 6 12 6

ˆ6 2 6 4

z

z

vl l

l l l lEI

vl ll

l l l l

θ

θ

− −

− − − −

11

1 1

2 2

22

ˆ

ˆ

ˆ

ˆ

QN

G MP

N Q

G M

− =

(3.28a)

Examining Figure 3.19a using Equation (3.28a), the internal forces in the beam for the

simple model are given in Table 3.5. It should also be noted that the external pressure

[fext] in Table 3.5 has already been computed in Equation (3.22).

Table 3.5: Internal forces using Equation (3.28a) for simple model

Ele

me

nt

1

Stiffness matrix Nodal Disp External Pressure Internal Forces

409092 1278413 -409092 1278413 0.00E+00 -36.66 0.00 -36.66

1278413 5326720 -1278413 2663360 2.45E-04 0.00 0.00 0.00

-409092 -1278413 409092 -1278413 1.35E-03 36.66 0.00 36.66

1278413 2663360 -1278413 5326720 1.59E-04 -229.13 0.00 -229.13

Ele

me

nt

2


409092 1278413 -409092 1278413 1.35E-03 -2.83 33.84 -36.66

1278413 5326720 -1278413 2663360 1.59E-04 229.13 0.00 229.13

-409092 -1278413 409092 -1278413 1.79E-03 2.83 22.56 -19.74

1278413 2663360 -1278413 5326720 -1.99E-05 -246.75 0.00 -246.75

Ele

me

nt

3


409092 1278413 -409092 1278413 1.79E-03 19.74 0.00 19.74

1278413 5326720 -1278413 2663360 -1.99E-05 246.75 0.00 246.75

-409092 -1278413 409092 -1278413 1.19E-03 -19.74 0.00 -19.74

1278413 2663360 -1278413 5326720 -1.59E-04 -123.37 0.00 -123.27

Ele

me

nt

4


409092 1278413 -409092 1278413 1.19E-03 19.74 0.00 19.74

1278413 5326720 -1278413 2663360 -1.59E-04 123.37 0.00 123.37

-409092 -1278413 409092 -1278413 0.00E+00 -19.74 0.00 -19.74

1278413 2663360 -1278413 5326720 -2.05E-04 0.00 0.00 0.00

In order to correctly interpret Table 3.5, one should recall the sign convection adopted

for positive shear force and bending moment in Equation (B.1) as follows:

( ) ( )1 2ˆ ˆ0 ; Q Q Q Q l= − = + (3.29a)

( ) ( )1 2ˆ ˆ0 ; M M M M l= − = + (3.29b)

[k] [u] [ku] [fext]

[fint]

[k] [u] [fext]

[fint]


71

Thus, examining Table 3.5 with Equation (3.29), one can conclude that the negative

values at local node 1 of the element (first two rows of each element) indicate a

positive internal force, while the negative values at local node 2 of element (last two

rows of each element) indicate a negative internal force. Similarly, examining Figure

3.19b using Equation (3.28b), the internal forces in the beam for the exact model are

presented in Table 3.6. Again, the external pressure [fext] is computed in Equation (3.22).

Table 3.6: Internal forces using Equation (3.28b) for exact model

Ele

me

nt

1


409092 1278413 -409092 1278413 0.00E+00 -36.66 0.00 -36.66

1278413 5326720 -1278413 2663360 2.65E-04 0.00 0.00 0.00

-409092 -1278413 409092 -1278413 1.48E-03 36.66 0.00 36.66

1278413 2663360 -1278413 5326720 1.79E-04 -229.13 0.00 -229.13

Ele

me

nt

2


409092 1278413 -409092 1278413 1.48E-03 -0.10 36.55 -36.66

1278413 5326720 -1278413 2663360 1.79E-04 279.89 50.76 229.13

-409092 -1278413 409092 -1278413 1.94E-03 0.10 19.85 -19.74

1278413 2663360 -1278413 5326720 -3.15E-05 -280.59 -33.84 -246.75

Ele

me

nt

3


409092 1278413 -409092 1278413 1.94E-03 19.74 0.00 19.74

1278413 5326720 -1278413 2663360 -3.15E-05 246.75 0.00 246.75

-409092 -1278413 409092 -1278413 1.26E-03 -19.74 0.00 -19.74

1278413 2663360 -1278413 5326720 -1.70E-04 -123.37 0.00 -123.27

Ele

me

nt

4


409092 1278413 -409092 1278413 1.26E-03 19.74 0.00 19.74

1278413 5326720 -1278413 2663360 -1.70E-04 123.37 0.00 123.37

-409092 -1278413 409092 -1278413 0.00E+00 -19.74 0.00 -19.74

1278413 2663360 -1278413 5326720 -2.17E-04 0.00 0.00 0.00

As before, one requires Equation (3.29) in order to correctly interpret Table 3.6.

Comparing the internal forces in Table 3.5 and 3.6 with Figures 3.20d and 3.21d, the

reader can see that the simple and exact models have the same internal forces as the

pressure applied between the two consecutive nodes. Moreover, the internal forces

computed by element stiffness [k] and its nodal displacement [u] given by [ku] are

identical to the ANSYS internal forces in Tables 3.1 to 3.4, which indicate that

ANSYS neglects the external pressures on an element when it computes the beam’s

internal forces, for a point force applied to a node on the beam. To summarize, the

[k]

[u]

[ku]

[fext] [fint]


72

author’s simple and exact numerical models apply an external pressure load to the

element. ANSYS then transfers these loads to the nodes as nodal forces. However,

ANSYS uses the element stiffness and nodal displacement to compute the internal

forces while neglecting the pressure load applied to the element. This, in turn, gives

an inaccurate solution as ANSYS does not adjust the internal forces to the correct

values. With the aid of Table 3.5 and Table 3.6, the author has shown how to correct

the internal forces. Nevertheless, the author is satisfied with the inaccurate numerical

values given by ANSYS.


73

3.2.3 Application of single load traversing Boyne Viaduct

The author is now in a position to analyse the Boyne Viaduct subjected to a single

moving load using the simple model. This system is used because of its faster

execution time and its reasonably good results, provided that the distance between

nodes is not excessively large. As described in Section 2.3, the author is only

concerned with the analysis of the centre span of the bridge; therefore, the outer

bridge structures are omitted from this study. However, approach spans with a zero

deflection are used, so the train experiences no vertical acceleration until it enters the

bridge. In the following example, the author focuses on a two-dimensional

representation of the Boyne Viaduct railway bridge as illustrated in Figure 3.24;

where all dimensions, section sizes and properties related to the bridge structure are

defined in Appendix F. As shown in Figure 3.24, the total length of the bridge is

80.77 m, with each of the 10 bays having a length 8.077 m.

Figure 3.24: Two-dimensional representation of the Boyne Viaduct railway bridge

The author begins by investigating the dynamic response of the two-dimensional

Boyne Viaduct with two different vehicle speeds: slow (10 km/hr or 2.778 m/s); and

fast (164 km/hr or 45.55 m/s). Furthermore, the weight of the moving load P is equal

to the weight of the bridge i.e. 1P G = . It is also worth noting that the faster speed of

164 km/hr is currently the maximum operating speed of the 201 Class locomotive

used by Irish-Rail (Wikipedia, 2007), while future locomotives purchased by Irish-

Rail are likely to reach speeds of 200 km/hr or higher.


74

In Figure 3.25 the vertical displacement and internal axial forces at mid-span of the

bridge as a function of time are plotted, respectively. In each case, the static effects

have been omitted from the results; thus, only the dynamic effects of the vehicle on

the bridge are observed and time t is arranged in such a manner that that the moving

load exits the right support of the bridge at 1ct L = . These results are then made

dimensionless by dividing the dynamic deflection and axial force, by the maximum

static deflection and axial force of the loaded structure at mid-span, and will be known as

the deflection coefficient and axial force coefficient, respectively. It can be seen from Figure

3.25 that the magnitude of the results is greatly influenced by the speed of the vehicle.

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.2 0.4 0.6 0.8 1.0

Dimensionless time ct /L

De

fle

cti

on

co

eff

icie

nt Moving load at 10 km/hr

Moving load at 164 km/hr

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

0.0 0.2 0.4 0.6 0.8 1.0


Ax

ial fo

rce

co

eff

icie

nt



Figure 3.25: (a) Deflection; (b) axial force at mid-span of the Boyne Viaduct subject

to a moving load travelling at 10 and 164 km/hr

(a)

(b)

Top chord

Bottom chord


75

Hence, a more in-depth analysis of the Boyne Viaduct subjected to a moving load at a

wider range of vehicle speeds is carried out. This varied speed can be made

dimensionless by means of the critical speed cr

c of the vehicle. The critical speed is

the speed of the vehicle at which the vehicle travels a distance of twice the length of

the bridge in a time equal to the first natural period of the bridge and is expressed

using Equation (C.53) as follows:

12cr

c f L= (3.30)

where 1f is the first natural frequency of the bridge. Therefore, the dimensionless

speed ratio α is then defined from Equation (C.52) as:

1

= 2

cr

c c

c f Lα = (3.31)

To explain Equations (3.30) and (3.31), the author must first conduct a modal analysis

of the two-dimensional Boyne Viaduct in order to determine its first natural

frequency. In Figure 3.26, the author presents the first two-flexural mode shapes and

their respective bridge frequencies obtained from a modal analysis. Using Equation

(3.30) and 1 3.403 Hzf = , the critical speed of a vehicle is computed as 550 m/s,=cr

c

which is equivalent to 1980 km/hr. Since the Boyne Viaduct is unlikely to ever

experience such a high speed, the author is really only interested in a more realistic

range of speeds between 0 km/hr to 300 km/hr i.e. 0 0.15α< < . However, given that

the results are presented in a dimensionless format, they can be compared with the


76

results of similar literature studies; thus, one also considers the ranges of vehicle

speeds between 0 1.0α< < .

Figure 3.26: 2D Natural frequencies and their corresponding mode shapes

In addition, the maximum dynamic deflection and axial force experienced by the

bridge at different speeds can be made dimensionless by dividing their value by the

maximum static load a mid-span; thus, defining their dynamic amplification factor

and are represented as follows (see Appendix G.3 for the vehicle positioning):

maximum dynamic deflection at midspan of the bridge =

maximum static deflection at midspan of the loaded bridgeU

DAF (3.32a)

maximum dynamic axial force at mid-span of the bridge =

maximum static axial force at mid-span of the loaded bridgeA

DAF (3.32b)

Figure 3.27 plots the dynamic amplification factor DAFU and DAFA, respectively, for

the range of speed values between 0 1.0α< < . Referring to Figure 2.2, the reader

should notice a favourable likeness between the dynamic amplification factor DAFU

of the Boyne Viaduct and a simply supported beam subjected to a moving load. From

the inspection of Figure 3.27, it is observed that the dynamic amplification factor

DAFU and DAFA are close to the minimum values at a speed ratio 0.21α ≈ , while at a

maximum value at a speed ratio 0.5α ≈ . However, these values fall outside the

realistic speeds, which are represented to the left of the vertical dashed line. A close-

up view of DAFU at these realistic speed values can be seen in Figure 3.28.

(a) 1st mode shape - 1 3.403 Hzf = (b) 2

nd mode shape - 2 9.872 Hzf =


77

1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

0.0 0.2 0.4 0.6 0.8 1.0

Speed ratio αααα

DA

FU

1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

0.0 0.2 0.4 0.6 0.8 1.0


DA

FA

Bot Chord

Top Chord

Figure 3.27: Dynamic amplification factor versus speed ratio at mid-span of the

Boyne Viaduct: (a) vertical displacement; (b) axial force

1.00

1.05

1.10

1.15

1.20

0 50 100 150 200 250 300

Speed (km/hr)

DA

FU

Figure 3.28: Close-up of Figure 3.27a with realistic vehicle speeds

(a)

(b)

Maximum realistic speed of a train traversing

the Boyne Viaduct (300 km/hr)


78

3.3 Multiple Moving Forces as a Function of Time

3.3.1 Development of multiple moving forces

The author expands the single moving force to several moving forces traversing a

beam. Originally, to avoid having more than one point force on an element at a time,

the length of a beam element had to be less than half the distance between wheels of a

vehicle. The drawback to this is that a large number of nodes are required to discretize

the beam. Therefore an alternative system, which allows several point forces on a

single finite element at once, is now developed. As before, the ANSYS node-to-

surface contact element system is also modified to include several moving forces for

comparison purposes. Equation (3.1b) is rewritten for several moving point forces as

follows:

( ) ( )2 2

, ,

carriages = bogie = wheels

,cN

j k m

j k m

p x t P x ctδ=

= − + ∆∑ ∑ ∑ , ,

0j k m

Lt

c

+ ∆≤ ≤ (3.33)

where ∆ incorporates the distance between carriages, the distance between bogies of a

single carriage, and the distance between wheels of a single bogie, which is only valid

when one or more moving point forces are on a particular beam element, L is the

length of the bridge.

3.3.1.1 Simple solution without overlapping time functions

Figure 3.29 presents a set of railway carriages that consists of a carriage supported by

two bogies, with each bogie supported by two axles, and a pair of wheels supports

each axle, which in the 2-dimensional model become a single point force. The front

pair of wheels of the carriage j is represented by the symbol Pj, the next pair of wheels

by Qj and the last two pairs by Rj and Sj respectively. According to Bowe &


79

Mullarkey (2000: p. 44), Aw is the distance the between the two axles of a single

bogie, Bw is the distance between centres of two bogies of a single carriage and Cw is

the distance between the rear axle of one carriage and the front axle of the carriage

behind it. It is clear from this set of dimensions that in the case of a single carriage the

distance between the rear axle of the front bogie and the front axle of the rear bogie is

(Bw-Aw). In this thesis, the distance Aw is considered to be smaller than the distance Cw

and (Bw-Aw). The force on each axle is P, which is a quarter of the combined weight of

a single carriage, two bogies and their associate axles and wheels.

Leading Carriage

Figure 3.29: Set of railway carriages (Bowe & Mullarkey, 2000)

From Figure 3.29, the distance from the front axle of the leading carriage to the front

axle of carriage j is as follows:

( )( ) ( )1 1 1j w w w w

P P A B C j D j= + + − = − (3.34)

The distance from the front axle of the leading carriage to the second axle of carriage j

is defined as:

( )( ) ( )1 1 1j w w w w w w

Q P A B C j A D j A= + + − + = − + (3.35)

Carriage jCarriage 1j +

1jP + jP 1PjQ 1Q1R

jR 1SjS

wB

wA

wC

w wB A−

w w w wD A B C= + +

For a typical train

(see Section G.2 in Appendix G)

( )w w w wA C B A< < −

Carriage 1


80

The distance from the front axle of the leading carriage to the third axle of carriage j

can be written as:

( ) ( ) ( )11 1

j w w w w w wR P A B C j B D j B= + + − + = − + (3.36)

And finally, the distance from the front axle of the leading carriage to the fourth axle

of carriage j is as follows:

( ) ( ) ( )11 1

j w w w w w w w wS P A B C j A B D j A B= + + − + + = − + + (3.37)

Now combining Equation (3.33), (3.34), (3.35), (3.36) and (3.37) in the right-hand

side of Equation (3.1) gives the following equation for several point forces traversing

a simply supported beam:

4 2

4 2

( , ) ( , )v x t v x tEI m

x t

∂ ∂+ =

∂ ∂

[ ( ) ( ) ( )2 2

carriages = bogie = wheels

1 1 1cN

w w w

j k m

P x ct D j B k A mδ=

− + − + − + − =

∑ ∑ ∑

[ ] [ ]1

( 1) ( 1)cN

w w w

j

P x ct D j x ct D j Aδ δ=

− + − + − + − +∑

[ ] [ ]( 1) ( 1)w w w w wx ct D j B x ct D j A Bδ δ+ − + − + + − + − + + (3.38)

where Nc is the number of carriages and ( ) ( ) ( ), , is 1 1 1

j k m w w wD j B k A m∆ − + − + − ,

which is only valid when the wheels of the train are between x = 0 and , ,

.j k m

x L= + ∆

Once the final wheel has left the bridge, the system then undergoes free vibration.

wheel 1 = P wheel 2 = Q

bogie 1

wheel 1 = R wheel 2 = S

bogie 2


81

The position of the m-th wheel, of the k-th bogie, of the j-th carriage, of the train can

be written as follows:

( ) ( ) ( )1 1 1 0w w w

x ct D j B k A m− + − + − + − = (3.39)

Using Equation (3.39), one is able to compute the time that a particular wheel arrives

on a specific point on the beam. For example, the time a wheel arrives on the left hand

support of the beam i.e. x = 0, one gets:

( ) ( ) ( )1 1 1 0w w wct D j B k A m− + − + − + − =

( ) ( ) ( )1 1 1w w w

D j B k A mt

c

− + − + −= (3.40a)

Equally, the time when a wheel arrives on the right hand support of the beam, x = L;

thus, giving:

( ) ( ) ( )1 1 1 0w w wL ct D j B k A m− + − + − + − =

( ) ( ) ( )1 1 1w w w

L D j B k A mt

c

+ − + − + −= (3.40b)

Similarly, the time when a wheel arrives on node i, x = xi, gives:

( ) ( ) ( )1 1 1 0i w w wx ct D j B k A m− + − + − + − =

( ) ( ) ( )1 1 1i w w w

x D j B k A mt

c

+ − + − + −= (3.40b)

The front wheel of the front bogie ( )1, 1m k= = of carriage j, Pj, is between node i-1

and node i between the times

−+−

c

jD

c

x wi )1(1 to ( 1)i wx D j

c c

− +

, and between


82

node i and node i+1 between the times ( 1)i wx D j

c c

− +

to 1 ( 1)i wx D j

c c

+ − +

. Since

wheel P1 is over left hand support at time t = 0, P1 influences node i between the

times c

xi 1− andc

xi 1+ ; hence, Pj influences node i between the times

−+−

c

jD

c

x wi )1(1

and

−++

c

jD

c

x wi )1(1 . The wheels Qj, Rj, and Sj follow a similar pattern to wheel Pj.

The issue of time just discussed can be incorporated in Equation (3.38) using

Heaviside functions H as follows:

4 2

4 2

( , ) ( , )v x t v x tEI m

x t

∂ ∂+ =

∂ ∂

[ ]( ) ( )

1

1 1( 1)

cNw w

w

j

D j D jLP x ct D j t t

c c cδ

=

− − − + − ⋅ − − − −

∑ H H

[ ]( ) ( )1 1

( 1)w ww w

w w

D j D jA ALx ct D j A t t

c c c c cδ

− − + − + − + ⋅ − − − − − −

H H

[ ]( ) ( )1 1

( 1)w ww w

w w

D j D jB BLx ct D j B t t

c c c c cδ

− − + − + − + ⋅ − − − − − −

H H

[ ]( ) ( )1 1

( 1)w ww w w w

w w w

D j D jA B A BLx ct D j A B t t

c c c c cδ

− − + ++ − + − + + ⋅ − − − − − −

H H

(3.41a)

where

( )0 for < 0

1 for 0

t

t

=

≥H t (3. 41b)


83

Figure 3.30 is a plot of the nodal force versus time and it tells the reader the effects of

carriage j and the front wheel of carriage j+1 on node i. In Figure 3.30 time breaks or

gaps are required between time-force functions to eliminate overlaps, which simplify

the computations i.e. linear functions.

ijInfluence of carriage on node

Gap 1 Gap 2 Gap 3

[xi-1 +

Dw (j-1

)]/c

[(xi +

Dw (j-1

)]/c

[xi+

1 +D

w (j-1)]/c

[xi-1 +

Dw (j-1

)]/c + A

w /c

[xi +

Dw (j-1

)]/c + A

w /c

[xi+

1 +D

w (j-1)]/c +

Aw /c

[xi-1 +

Dw (j-1

)]/c + B

w /c

[xi +

Dw (j-1

)]/c + B

w /c

[xi+

1 +D

w (j-1)]/c +

Bw /c

[xi-1 +

Dw (j-1

)]/c + (A

w +

Bw )/c

[xi +

Dw (j-1

)]/c + (A

w +

Bw )/c

[xi+

1 +D

w (j-1)]/c +

(Aw

+ B

w )/c

[xi-1 +

Dw ( j )]/c

Figure 3.30: Influence of carriage j on node i (Bowe & Mullarkey, 2000)

With the aid of Figure 3.30, it is clear that the size of Gap 1 is as follows:

[ ] [ ])1(1

)1(1

11 −+−+−+ +− jDxcc

AjDx

cwi

wwi = [ ]

c

Axx

c

wii +− +− 11

1 =

c

A

c

l w+−2

= c

lAw 2− (3.42)

Gap 1 must be greater than zero to prevent any overlap; therefore, Aw must be the

greater than 2l, where l is distance between nodes on the beam. This implies that there

should be at least two elements between the two axles of a bogie. Similarly from

Figure 3.30, Gap 2 is defined as:

t

Wheel jP Wheel jQ Wheel jR Wheel jS 1Wheel jP +

P


84

[ ]c

ABxx

c

ww

ii

−+− +− 11

1 =

c

AB

c

l ww −+

−2 >

c

A

c

l w+− 2

, because (Bw - Aw) > Aw (3.43)

Therefore Gap 2 is greater than zero. Using Figure 3.30, Gap 3 can be written as:

[ ]c

BA

c

Dxx

c

www

ii

)(111

+−+− +− =

c

C

c

l w+− 2

> c

A

c

l w+− 2

> 0, because Cw > Aw (3.44)

Again, Gap 3 is clearly greater than zero because Gap 1 is greater than zero.

The inequality Aw > 2l demands that a large number of nodes be used to discretize the

beam; however, this model is less complicated than other techniques and still has a

reasonably quick execution time. Getting rid of Gaps 1, 2 and 3 has the advantage of

reducing the number of nodes required on the beam; however, the complexity of the

force as a function of time greatly increases as the number of wheels on a beam

element increases.

3.3.1.2 Simple and exact models with overlapping time functions

In the previous section, the notation of overlapping time function is not considered as

the distance between wheels of a single bogie Aw is always larger than twice the

length of a single element; however, the impracticality of that model is that as the

distance between wheels decreases, the number of elements required in the model

increases. Therefore, by considering overlapping time function, one can reduce the

number of elements in the model as the distance between axles of a single bogie Aw,

no longer determines the length of a beam element l. Nevertheless, the drawback to

this system is that the overlapping time-function requires a larger preparation time, in

the pre-processing step, in order to compute the more complex time function.

Fortunately, Equation (3.38) can still be used to calculate this complex overlapping

time function.


85

For demonstration purposes, Figure 3.31 presents the spatial and time domain of two

moving forces, with the same magnitude, traversing a beam; for several values of

wA l for node i.

(a) Influence of two point forces at a distance 2l apart on node i

(b) Influence of two point forces at a distance 1.5l apart on node i

(c) Influence of two point forces at a distance l apart on node i

Spatial Domain for node i Time Domain for node i

2wA l=

1Q1P

l

i

P

4l

c

t0

7

2

l

c

t

P

1Q 1P

1.5wA l=

l

i

3l

cl

wA l=

1Q1P

i t

P

0

0

(1)

(2)

(3)

(4)

(5)

(6)

(1)

(2)

(3)

(5)

(6)

(4)

(1)

(2)

(6)

(3) (4)

(5)


86

(d) Influence of two point forces at a distance 0.5l apart on node i

Figure 3.31: Relationship between spatial and time domain under different load

conditions for two moving forces

Examining Figure 3.31, the first observation is that the time lag between the two loads

is w

A c ; therefore, each time domain occurs over 2

wA l

c

+. The next observation is

that when the first load P1 is on the element to right of node i, while the second load

Q1 is on the element to left of node i, the magnitude of the load applied is constant as

illustrated in Figure 3.31(b) to 3.31(d). The final observation is that when the two

loads are on the same element to the left or right of node i, the slope of the time

function doubles as shown in Figure 3.31(d). For completeness, the author also plots

the time domains, of the two moving forces using the exact numerical model with

overlapping time functions (forces and moment) and can be found in Figure G.6 in

Appendix G.

t0

(1) P1 arrives on node i-1; (2) P1 arrives on node i; (3) P1 arrives on node i+1

(4) Q1 arrives on node i-1; (5) Q1 arrives on node i; (6) Q1 arrives on node i+1

(6) 5

2

l

cl

0.5wA l=P

i

1Q 1P

(1)

(2)

(3)

(5)

(4)


87

3.3.1.3 Multiple moving forces using ANSYS contact elements

Modifying Section 3.2.1.3, the author expands the system to incorporate several

railway carriage wheels, which in turn increases the number of contact elements

required in this analysis. The front pair of wheels of carriage j is represented by the

point mass element MP j, the next pair of wheels by MQ

j and the last two pairs by MR

j

and MS j respectively as can be seen in Figure 3.32 (Bowe & Mullarkey, 2000; p. 47).

4 Contact Elements

Target Surface

Leading Carriage

PPPPPPP PP PPP

Carriage 1jCarriagej +1Carriage

Figure 3.32: Establishing contact elements between wheels and beam

The nodes associated with these point mass elements are once again defined as

contact nodes, while the nodes on the beam are defined as the target surface. Contact

elements are then established between the contact nodes and the target surface as

shown in Figure 3.32. Each point mass element is given zero mass and a vertical force

P is applied to the associated node.

It should be noted that using the ANSYS node-to-surface contact elements to model

several moving forces has its drawbacks, as the user must also provide entry and exit

approaches on either side of the beam to simulate the train entering and leaving the

bridge structure; thus, increasing the number of contact elements required.

1j

SM + 1j

RM + 1j

QM + 1j

PM + 1

SM 1

RM1

QM 1

PM

wB wC wAw wB A−

w w w wD A B C= + +

j

SM j

RMj

QM j

PM


88

3.3.2 Validation of multiple moving forces as a function of time

In order to validate the author’s multiple moving force system; one again compares

the results from the developed systems with the results of the contact elements of

ANSYS using an example obtained from the literature.

In this first example, the bridge and train properties adopted are similar to those of

Yang & Wu (2001), such that the bridge has a length L of 30m, Young’s modulus of

elasticity E of 2.87x107 kN/m



m of 32.4 t/m, Poisson’s ratio ν of 0.2; hence, the first natural frequency of the bridge

ω1 is 25.666 rad/sec. Again, one ignores the gravitational and damping effects of the

bridge. In contrast, the train comprises of 10 identical railway carriages traversing the

bridge at a constant speed c of 27.778 m/s (100 km/hr), with each railway carriage

consisting of a two-wheel assembly i.e. no bogies. The distance between wheels of a

single carriage Aw is 19 m, while the distance between the rear wheel of one carriage

and the front wheel of the following carriage Cw is 6m; therefore the repetitive

distance between the front wheels of each carriage Dw = 19 + 6 = 25 m. Each wheel of

the railway carriage applies a force P of -218.2 kN to the bridge. As before, the

Newmark-β time integration method (Bathe, 1996) with 1000 equal time steps is used

to solve the transient analysis. Time t is arranged in such a manner that the front

wheel of the first railway carriage is at the left hand support at t = 0 sec. In addition,

the initial displacement and velocity of the beam are equal to zero at time t = 0 sec.

For the simple model without overlapping time functions the bridge has been

discretized into 16 beam elements, while only 8 beam elements are used to discretize

the bridge for the simple model with overlapping time functions as well as when using

the ANSYS contact element. All approach spans used have zero deflections.


89

Figure 3.33 plots the influence of four wheels i.e. the 2nd

to the 5th

wheel of the train,

on the mid-span node of the bridge as a function of time for the simple model with

and without overlapping time function. One can see from Figure 3.33 that the simple

model with overlapping time function influences the mid-span node over a longer

time interval because there are fewer nodes used in that system.

-300

-250

-200

-150

-100

-50

0

1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6

Time (sec)

Fo

rce

(k

N)

Simple Model - No Overlapping

Simple Model - Overlapping

Figure 3.33: Influence of the 2nd

to the 5th

wheel of the train on a mid-span node

The vertical displacement, acceleration and bending moment at mid-span of the beam

as a function of time for the three different systems can be seen in Figure 3.34. It can

be seen that there is a great likeness among the results of all three systems in

magnitude and curvature. It can also be noted that the leading wheel of the train

arrives at the right support at 30/27.778 = 1.08 sec, as indicated by the vertical dashed

line on each diagram.

No Overlapping

(1) 2nd wheel left of mid-span node; (2) 2nd wheel at mid-span node; (3) 2nd wheel right of mid-span node

(4) 3rd wheel left of mid-span node; (5) 3rd wheel at mid-span node; (6) 3rd wheel right of mid-span node

Overlapping

(i) 2nd wheel left of mid-span node; (ii) 2nd wheel at mid-span node; (iii) 2nd wheel right of mid-span node

(iv) 3rd wheel left of mid-span node; (v) 3rd wheel at mid-span node; (vi) 3rd wheel right of mid-span node

(i) (1) (3) (4) (6) (vi)

(2)

(ii) (5)

(v)

(iv) (iii)


90

-0.0015

-0.0012

-0.0009

-0.0006

-0.0003

0.0000

0 2 4 6 8 10

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

) Simple Model - No Overlapping


ANSYS CE

-0.30

-0.20

-0.10

0.00

0.10

0.20

0 2 4 6 8 10

Time (sec)

Ve

rtic

al A

cc

ele

rati

on

(m

/s2)



ANSYS CE

0

1000

2000

3000

4000

5000

0 2 4 6 8 10

Time (sec)

Be

nd

ing

Mo

me

nt

(kN

m)



ANSYS CE

Figure 3.34: (a) Vertical displacement; (b) acceleration; (c) bending moment at

mid-span of the beam as a function of time for 10 identical railway

vehicles

Leading wheel over RHS

(c)

(b)

(a)


91

The second example considered uses the same bridge and train properties to those of

Yang et al. (1997), such that the bridge has a length L of 20m, Young’s modulus of

elasticity E of 29.43x106 kN/m



m of 34.088 t/m, Poisson’s ratio ν of 0.2; therefore, the first natural frequency of this

bridge ω1 is 44.75 rad/s (or f1 = 7.122 Hz). As before, one ignores the gravitational

and damping effects of the bridge. The train model comprises of 5 railway carriages,

each having a car length Dw of 24m, with each railway carriage consisting of a two-

wheel assembly i.e. no bogies. The distance between wheels of a single carriage Aw is

18 m and the force applied to the bridge by each wheel P is –215.6 kN. Again, time t

is arranged in such a manner that the front wheel of the first railway carriage is at the

left hand support at t = 0 sec. In addition, the initial displacement and velocity of the

beam are equal to zero at time t = 0 sec. In this example, the bridge is discretized into

10 beam elements for the simple model with overlapping time functions.

The vertical displacement at mid-span of the beam as a function of time for two

different train speeds; 26 m/s (94 km/hr) and 34 m/s (122 km/hr) are plotted in

Figures 3.35a and 3.35b, respectively. From inspection of Figure 3.35a, the reader can

see that with each passing vehicle; the bridge tends to deflect by a similar amount

each time, while in Figure 3.35b the bridge is inclined to experience an increase in the

deflection similar to the effects of resonance with each passing vehicle. Furthermore,

with a change of notation, Yang et al. (1997) and Fryba (2001) formulise that the

resonance of bridges is likely to occur due to the repetitive vehicle loading of the front

wheel of the train at the following speed ratio:

1, 2, 3...2

wD

nnL

α = = (3.45)


92

Substituting Equations (3.30) and (3.31) into Equation (3.45), one then finds the speed

at which resonance is likely to occur on the bridge, as follows:

1

1, 2, 3...2 2

w

cr

Dc cn

c f L nL= = = (3.46a)

1 1, 2, 3...wf D

c nn

= = (3.46b)

In addition, Yang et al. (1997) has formulised a condition of cancellation, whereby

repetitive vehicle loadings can suppress the bridge resonance and can be written in

terms of the speed ratio, with a change of notation, as follows:

1 1, 2, 3...

2 1n

nα = =

− (3.47)

Substituting Equations (3.30) and (3.31) into Equation (3.47) then gives the speeds at

which the condition of cancellation is likely to occur on the bridge as follows:

1 1, 2, 3...

2 1cr

cn

c n= =

− (3.48a)

12

= 1, 2, 3...2 1 2 1

crc f L

c nn n

= =− −

(3.48b)

Using Equations (3.45) and (3.46b) as well as Equations (3.47) and (3.48b), the author

computes six speeds at which resonance and cancellation effects are likely to occur

for this particular bridge due to repetitive vehicle loading and these results are

presented in Table 3.7. From the results in Table 3.7, one can see that resonance of the

bridge occurs when the train crosses the bridge at 34.2 m/s (123.1 km/hr), which is in

good agreement with Figure 3.35b.


93

Table 3.7: Yang et al. (1997) bridge model - resonance and cancellation speeds

Moreover, from the results in Table 3.7 it can be seen that when the train has a speed

of 25.9 m/s (93.6 km/hr), the condition of cancellation occurs, which is in good

agreement with Figure 3.35a. As an example, to show that the condition of

cancellation suppresses the risk of resonance on the bridge, the author will also plot

the vertical displacement at mid-span of the beam as a function of time subjected to a

train travelling at 57 m/s (205 km/hr), which is presented in Figure 3.35c. From

inspection of Table 3.7, the reader can see that both resonance and cancellation share

this particular speed.

-1.00

-0.75

-0.50

-0.25

0.00

0.25

0.50

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

m)

Simple model - Overlapping - Train speed = 26m/s

Yang et al. (1997) Solution

Risk of Resonance Condition of Cancellation

Eq. (3.45) Eq. (3.46b) Eq. (3.47) Eq. (3.48b)

n α c (m/s) α c (m/s)

1 0.60 170.9 1.00 284.9

2 0.30 85.5 0.33 95.0

3 0.20 57.0 0.20 57.0

4 0.15 42.7 0.14 40.7

5 0.12 34.2 0.11 31.7

6 0.10 28.5 0.09 25.9


(a)


94

-1.00

-0.75

-0.50

-0.25

0.00

0.25

0.50

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

m)

Simple model - Overlapping - Train speed = 34m/s


-1.00

-0.75

-0.50

-0.25

0.00

0.25

0.50

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5

Ve

rtic

al D

isp

lac

me

nt (m

m)

Dimensionless time ct/L

Simpel model - Overlapping - Train speed = 26 m/s



Figure 3.35: Vertical displacement at mid-span of the beam: (a) Train speeds of

26m/s; (b) Train speeds of 34m/s; (c) Train speeds of 57m/s

As can be seen in Figure 3.35c, the vertical displacement at mid-span of the beam

tends to deflect by the same amount with each passing train. This indicates that

resonance does not occur at this particular speed, which is in good agreement with

Yang et al. (1997) that the cancellation effect supersedes the resonance effects. It can

be noted that the first natural frequency of long spanning bridges is generally low,

which makes them more susceptible to resonances of carriages with closely spaced axles.


(b)

(c) Rear wheel of last carriage on RHS


95

One finishes by examining the dynamic deflection of this particular bridge at various

train speeds using the simple model with overlapping time functions. From the

literature as well as Yang et al. (1997) the impact factor IF can be defined as follows:

( ) ( ) ( )1 1

( ) ( )

d s d

U

s s

R x R x R xIF DAF

R x R x

−= = − = − (3.49)

whereby Rd is the maximum dynamic deflection of beam and Rs is the maximum

static deflection of the beam at position x due to the action of the moving loads. In this

particular example, the maximum deflection will occur at mid-span of the beam.

A plot of the impact factor IF for a range of train speed ratios between 0 1.0α< < is

shown in Figure 3.36. In this diagram, the reader can see a good similarity between

the simple model with overlapping time functions and the numerical solution of Yang

et al. (1997). In addition, one can see that the largest impact factor of 5.5,IF = occurs

when the speed ratio α is equal to 0.6, which coincides with Table 3.7.

0.0

1.0

2.0

3.0

4.0

5.0

6.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8


Imp

ac

t fa

cto

r I



Figure 3.36: Impact factor at mid-span of the beam at various train speeds


96

3.3.3 Application of multiple moving forces traversing the Boyne Viaduct

In this section, the author examines the dynamic response resulting from several

moving loads traversing the Boyne Viaduct railway bridge, where the bridge is

represented as a two-dimensional or three-dimensional structure. In addition, a twin

track structure subjected to a single train or a pair of trains moving in opposite

directions can be easily implemented. In each example, a typical Irish-Rail vehicle is

chosen.

However, to the author’s knowledge, there has never been any experimental data

collected from this particular bridge; hence, the numerical solution obtained in this

thesis is merely a demonstration of the developed systems acting on a real structure.

Nonetheless, Sections 3.2.2 and 3.3.2 have shown that the results of these developed

moving load systems compare admirably with the results of other research studies.

3.3.3.1 Railway Vehicles

The railway vehicles adopted in each of the following examples are a six-axle 201

class locomotive and a four-axle Mark 3 railway coach. The axle spacing and weights

of these particular vehicles were supplied by the Irish Rail’s Structural Design Office,

Inchicore, Dublin and are presented in Figure 3.37. As shown in Figure 3.37, each

wheel of the locomotive exerts a force of 91.25 kN on the rail and each wheel of the

railway coach exerts a force of 58.86 kN on the rail, respectively. In order for the

Boyne Viaduct to experience the full weight of the train over its entire bridge length,

the author has chosen a train consisting of a single six-axle 201 Class locomotive (see

Figure 3.37a) pulling three four-axle Mark 3 railway coaches (see Figure 3.37b), such

that the total distance between the front axle and rear axle of the train is equal to

( ) ( )21.051 3 23.000 1.969 2.200 85.882m.+ × − + =


97

Figure 3.37: Typical axle spacing and load of Irish-Rail vehicles: (a) 201 Class

locomotive; (b) Mark 3 railway coach

3.3.3.2 Two-Dimensional Boyne Bridge

Modifying Section 3.2.3, the two-dimensional Boyne Bridge is now subjected to a

moving train comprising a locomotive and several railway carriages, which can be

simulated using either the simple model with or without overlapping time functions.

However, it is important to note that for the simple model without overlapping time

function, one requires that the bottom chord of the truss be discretized into 100 beam

elements i.e. 10 beam elements per bay, whereas with overlapping time functions, one

can use a single beam element per bay; thus, only requiring 10 beam elements in the

bottom chord of the truss. Table 3.8 presents the time of execution for the simple

model with and without overlapping time function as well as for the contact element

system with 1000 equal time-steps using an Intel Core Duo 2.13 GHz processor with

2048 MB RAM.

(a)

(b)

Mark 3 Railway Coach

201 Class locomotive

Gross Weight = 48 tons

Axle Weight = 12 tons

Axle Load = 117.72 kN

Wheel Load = 58.86 kN

Gross Weight = 111.5 tons

Axle Weight = 18.6 tons




98

Table 3.8: Time of execution for the developed systems

Time of execution in sec

Timesteps

Simple Model ANSYSCE

No Overlapping Overlapping

(100 beam elements) (10 beam elements) (10 beam elements)

1000 0 42* 0

26 60 55

* = precalculation time prior to executing model

It can be seen from Table 3.8 that the time of execution for the simple model without

overlapping is over 100% faster than the other two systems. However, much of the

slowness for the overlapping system is due to the pre-calculations of the overlapping

time-force functions before it executes the model. By omitting this pre-calculation

from its total execution time, it was found that the overlapping system is

approximately 33% faster than the former method. Nonetheless, most results obtained

in this section use the simple model without overlapping due to its speed of execution.

Figure 3.38 plots the influence of the train on the mid-span node of the bridge as a

function of time for the simple model with and without overlapping.

-300

-250

-200

-150

-100

-50

0

0.4 0.6 0.8 1 1.2 1.4 1.6 1.8


Fo

rce

(k

N)



Figure 3.38: Influence of the train on a mid-span node


99

Somewhat similar to Section 3.2.3, the author examines the dynamic effects of a train

traversing the Boyne Viaduct at a slow (10km/hr) and fast speed (164 km/hr). In

Figures 3.39 the reader can see a plot of the vertical displacement and axial force in

the top and bottom chord at mid-span of the bridge as a function of time. It should

also be noted from the diagram that the front wheel of the train exits the bridge

structure at time 1ct L = . In Figure 3.39, one can see that the dynamic effects are

minimal at the slow speed, while at the higher speed the bridge tends to experience a

noticeable oscillation.

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0


De

fle

cti

on

co

eff

icie

nt

Multiple Moving Loads - 10 km/hr


-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

0.0 0.5 1.0 1.5 2.0


Ax

ial fo

rce

co

eff

icie

nt



Figure 3.39: (a) Vertical displacement; (b) axial force at mid-span of the 2D Boyne

Viaduct due to a train travelling at 10 km/hr and 164 km/hr

(a)

(b)

Bottom Chord

Top Chord


100

In addition, plots of the dynamic amplification factors DAFU and DAFA for the range

of speed values between 0 1.0α< < can be seen in Figure 3.40. Unlike in Figure

3.27, the maximum dynamic deflection experienced by the bridge tends to occur at the

critical speed of the vehicle i.e. α = 1.

1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

0.0 0.2 0.4 0.6 0.8 1.0


DA

FU

201 Loco + 3 MK3 coaches

1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

0.0 0.2 0.4 0.6 0.8 1.0


DA

FA

201 Loco + 3 MK3 coaches (Top Chord)

201 Loco + 3 MK3 coaches (Bottom Chord)

Figure 3.40: Dynamic amplification factor at mid-span of the 2D Boyne Viaduct

versus speed ratio: (a) vertical displacement; (b) axial force

In Figure 3.41, the author presents a close-up view of the DAFU taken from Figure

3.40a; where the vertical dash line represents the maximum realistic vehicle speed that

(a)

(b)

Maximum realistic speed of a train traversing the

Boyne Viaduct (300 km/hr)


101

is of interest i.e. 0.15α = implies that 300 km/hr.c = Examining Figure 3.41, it can

be seen that the results for several moving loads tends to be much lower than the

results for single moving load, as described in Section 3.2.3, mainly because the

several moving loads are applied to the bridge over its length and not concentrated at

a single point. Nevertheless, the curvature of each set of results is very similar.

1.00

1.05

1.10

1.15

1.20

0 50 100 150 200 250 300

Speed (km/hr)

DA

FU

Multiple Moving Loads

Single Moving Load - Figure 3.28


3.3.3.3 Three-Dimensional Boyne Bridge

Until now, the author has only considered a two-dimensional representation of the

Boyne Viaduct railway bridge whereas a three-dimensional bridge would better

represent the actual structure. Therefore a comparison is made between the results of

the two-dimensional and three-dimensional Boyne Bridge subjected to multiple

moving loads using either of the developed moving load systems.

Figure 3.42 presents a three-dimensional model of the Boyne Bridge, developed using

the ANSYS finite element program. Section sizes and properties related to the bridge

structure can be found in Appendix F. Unlike the two-dimensional structure, cross-


102

beams are located between the main nodes of the truss, which in turn support the

longitudinal beams of the three-dimensional bridge structure. The railway tracks are

then positioned directly above the longitudinal beam; however, the railway tracks are

omitted in this study. Nevertheless, one does simulate the train traversing the bridge

along these longitudinal beams. Delgado & Dos Santos (1997) paper presents a train

traversing a bridge at a wide range of speeds with and without bridge ballast. From

their study it is shown that the bridge ballast does not influence their results hence, the

bridge ballast on the Boyne Viaduct has been omitted from this study.

Figure 3.42: Three-dimensional representation of the Boyne Bridge

In the following example, the three-dimensional Boyne Bridge is subjected to the

same train loads used in Section 3.3.3.2 on each rail, while travelling at a slow (10

km/hr) and fast speed (164 km/hr). In Figures 3.43a and 3.43b, the author plots the

vertical displacement and axial forces in the top and bottom chord at mid-span of the

bridge as a function of time. Like the previous results of Figure 3.39, the dynamic

response tends to be minimal at the slow speed, while at the higher speed the bridge

tends to experience slightly less oscillation than that of its two-dimensional

counterpart. In Figures 3.43c the bending moment at mid-span of the cross-beam

located at mid-span of the bridge as a function of time for both speeds is presented. It

should be noted from the diagrams that the front wheel of the train leaves the bridge at

1.ct L =

Longitudinal beams

Cross-beams


103

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0

Dimesionless time ct /L

De

fle

cti

on

co

eff

icie

nt



-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

0.0 0.5 1.0 1.5 2.0


Ax

ial fo

rce

co

eff

icie

nt



0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0


Be

nd

ing

mo

me

nt

co

eff

icie

nt



Figure 3.43: (a) Vertical displacement; (b) axial force (c) bending moment of the

cross-beam located at mid-span of the 3D Boyne Viaduct due to a train

travelling at 10 km/hr and 164 km/hr

(a)

(b)

(c)

Bottom Chord

Top Chord


104

Prior to examining the dynamic amplification factors DAFU and DAFA for the three-

dimensional structure, the author must now re-calculate the first natural frequency of

the three-dimensional Boyne Viaduct using a modal analysis, as this structure has

become a lot stiffer and heavier than its two-dimensional counterpart. The extra mass

and stiffness associated with the three-dimensional bridge model is due to the addition

of cross-beams and longitudinal beams, which is not modelled in the two-dimensional

bridge model. In Figure 3.44, the author presents the first two-flexural mode shapes

and their respective bridge frequencies obtained from a modal analysis. Using

Equation (3.30) and 1

=f 3.123 Hz , the critical speed of a vehicle is computed as

504 m/s,=cr

c which equivalent to 1814 km/hr. As before, the Boyne Viaduct is

unlikely to ever experience such a high speed; hence, a more realistic range of speeds

between 10 km/hr to 300 km/hr i.e. 0 0.165,α< < are of greatest concern in the

author’s parametric study.

Figure 3.44: 3D Natural frequencies and their corresponding mode shapes

A plot of the dynamic amplification factors DAFU and DAFA for the range of speed

values between 0 1.0α< < can be seen in Figure 3.45. From inspection of Figure

3.45, it can be seen that the dynamic amplification factors are similar for the two-

dimensional and three-dimensional bridge models at speeds 0.4,α < however, as the

speed ratio increases the DAFU and DAFA for the two-dimensional model and three-

dimensional vary by as much as 10%.

(a) 1st mode shape -

13.123 Hz=f (b) 2

nd mode shape -

28.716 Hzf =


105

1.0

1.2

1.4

1.6

1.8

2.0

0.0 0.2 0.4 0.6 0.8 1.0


DA

FU

Multiple Moving Loads - 2D Boyne


1.0

1.2

1.4

1.6

1.8

2.0

0.0 0.2 0.4 0.6 0.8 1.0


DA

FA

Multiple Moving Loads - 2D Boyne (Bot)

Multiple Moving Loads - 3D Boyne (Bot)

1.0

1.2

1.4

1.6

1.8

2.0

0.0 0.2 0.4 0.6 0.8 1.0


DA

FA

Multiple Moving Loads - 2D Boyne (Top)

Multiple Moving Loads - 3D Boyne (Top)


versus speed ratio: (a) vertical displacement; (b) axial force in the

bottom chord; (c) axial force in the top chord at a range of speeds

(b)

(a)

(c)


Boyne Viaduct (300 km/hr)


106

A close-up view of the DAFU obtained from Figure 3.45a can be seen in Figure 3.46,

which uses realistic vehicle speeds instead of the speed parameter; therefore,

0.165α = implies that 300 km/hr.c = Examining Figure 3.46, it can be seen that the

two-dimensional and three-dimensional results are fairly similar to each other apart

from a slight shift of results. This shift of results can be partially due to the difference

observed between the natural frequencies of the two-dimensional (see Figure 3.26)

and three-dimensional bridges (see Figure 3.44). The vehicle in the three-dimensional

model has a smaller critical speed (504 m/s) than the vehicle in the two-dimensional

model (550 m/s).

1.00

1.05

1.10

1.15

0 50 100 150 200 250 300

Speed (km/hr)

DA

FU




3.3.3.4 Twin-track Railway Bridge

Using the simple model with overlapping time functions, one can easily model two

trains travelling in opposite directions on a twin-track bridge. For convenience, one

widens the Boyne Viaduct railway bridge to allow space for an additional railway

track as shown in Figure 3.47.


107

Figure 3.47: Twin-track railway bridge model

Using the same section sizes and properties of Section 3.3.3.3, the author conducts a

modal analysis of this structure in order to compute its critical speed. Figure 3.48

presents the first two-flexural mode shapes and their respective bridge frequencies

obtained from a modal analysis. Using Equation (3.30) and the first natural frequency

of the bridge1 of 2.801 Hz,f the critical speed of a vehicle is computed as

452 m/s,=cr

c or 1627 km/hr. As in the previous section, it is unlikely that such a

high speed would ever be experienced on this bridge; thus, a more realistic range of

speeds between 10 km/hr to 300 km/hr i.e. 0 0.185,α< < are of interest.

Figure 3.48: Twin-track frequencies and their corresponding mode shapes

The author only examines two railway vehicles arriving on the bridge at the same

time travelling at various speeds in opposite directions as shown in Figure 3.49. In

each example, one assumes that both trains are identical in weight and length.

Twin track

(a) 1st mode shape -

12.801 Hz=f (b) 2

nd mode shape -

25.994 Hz=f


108

Figure 3.49: Bridge subjected to two trains arriving at the same time (Plan View)

Figure 3.50 presents the dynamic amplification factor DAFU at mid-span of the twin-

track structure, which is subjected to two trains traversing the bridge at a range of

speeds between 0 0.5α≤ ≤ . On this occasion, the static loaded deflection of the

bridge takes into account the weight of two parked trains on the bridge; therefore, it

can be seen from Figure 3.50 that the DAFU can have a value less than 1.0 when the

trains have significantly different speeds i.e. one train has a high speed while the other

has a low speed. From inspection of Figure 3.50, it can be seen that the maximum

deflection of the bridge occurs when the trains have a relatively similar speed, which

is also observed by (Lin & Ju, 2003: p. 106).

For a more in-depth analysis of a twin-track railway bridge, one can also vary the

vehicle weight, length and arrive time; however this goes beyond the scope of this

study. Nonetheless, the author’s developed systems are not limited to single track

structures.


109

Figure 3.50: Dynamic amplification factor DAFU at mid-span of the twin track

structure due to two trains traversing simultaneously

3.3.3.5 Boyne Viaduct subjected to Eurocodes (1991) rail traffic loads

This final section subjects the three-dimensional Boyne Viaduct to rail traffic loads

found in Eurocodes (EN 1991-2, 2003), in particular, Type 1 and Type 8 train loads.

The Type 1 train model consists of a locomotive and 12 railway coaches as shown in

Figure 3.51a, while the Type 8 train model comprises a locomotive and 20 freight

wagons as shown in Figure 3.51b. One analyses the Type 1 train model traversing the

Boyne Viaduct at 200 km/hr (the speed suggested in the Eurocodes). The Type 8 train

traverses the Boyne Viaduct at 100 km/hr (the speed suggested in the Eurocodes). In

each example, the simple model with overlapping is used.

DAFU


twin track bridge (300 km/hr)


110

Figure 3.51: Eurocode train models (EN 1991-2, 2003): (a) Type 1; (b) Type8

Figure 3.52 plots the influence of both train types acting on the mid-span node of the

bridge as a function of time for the simple model with overlapping. The vertical

displacement (Figure 3.53) and axial forces in the top chord (Figure 3.54) and bottom

chord (Figure 3.55) at mid-span of the Boyne Viaduct subjected to the two different

train types are shown below. For train Type 1, the maximum static deflection at mid-

span of the loaded bridge is 0.03295 m (32.95 mm), while the maximum static axial

force at mid-span of the loaded bridge are -1637 kN and 1057 kN in the top and

bottom chords of the bridge, respectively. Train Type 8 has a maximum static

deflection at mid-span of the loaded bridge of 0.04809 m (48.09 mm), while the

maximum static axial forces at mid-span of the loaded bridge are -2257 kN and 1500

kN in the top and bottom chords of the bridge, respectively.

From inspection of Figures 3.53 to 3.55, one can see that the dynamic coefficient for

the Boyne Viaduct tends to experience periodic oscillations for the Type 1 train, while

little to no response is observed for the Type 2 train, as both trains traverses the bridge

at the chosen speeds given in the Eurocodes (1991).

(a)

(b)

Chapter 3 – Wheel forces represented as time varying nodal forces and moments

111

-350

-300

-250

-200

-150

-100

-50

0

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0


Fo

rce

(kN

)

Figure 3.52a: Influence of train Type 1 on a mid-span node

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5


Defl

ec

tio

n c

oe

ffic

ien

t

Train Type 1 travelling at 55.55 m/s


Figure 3.53a: Vertical displacement at mid-span of the Boyne Viaduct subjected

to train Type 1 traversing at 55.55 m/s (200 km/hr)

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5


Ax

ial

forc

e c

oe

ffic

ien

t



Figure 3.54a: Axial force in the top chord at mid-span of the Boyne Viaduct

subjected to train Type 1 traversing at 55.55 m/s (200 km/hr)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5


Ax

ial

forc

e c

oe

ffic

ien

t



Figure 3.55a: Axial force in the bottom chord at mid-span of the Boyne Viaduct


-350

-300

-250

-200

-150

-100

-50

0

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5


Fo

rce

(k

N)

Figure 3.52b: Influence of train Type 8 on a node a mid-span

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0


De

fle

cti

on

co

eff

icie

nt



Figure 3.53b: Vertical displacement at mid-span of the Boyne Viaduct subjected to

train Type 8 traversing at 27.78 m/s (100 km/hr)

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0


Ax

ial

forc

e c

oe

ffic

ien

t



Figure 3.54b: Axial force in the top chord at mid-span of the Boyne Viaduct

subjected to train Type8 traversing at 27.78 m/s (100 km/hr)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0


Ax

ial

forc

e c

oe

ffic

ien

t



Figure 3.55b: Axial force in the bottom chord at mid-span of the Boyne Viaduct


Chapter 3 – Wheel forces represented as time varying nodal forces and moments

112

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5


De

flec

tio

n c

oeff

icie

nt



Figure 3.56a: Vertical displacement at mid-span of the Boyne Viaduct subjected

to train Type 1 traversing at 63.4 m/s (228.24 km/hr)

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5


Axia

l fo

rce

co

eff

icie

nt



Figure 3.57a: Axial force in the top chord at mid-span of the Boyne Viaduct

subjected to train Type 1 traversing at 63.4 m/s (228.24 km/hr))

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5


Axia

l fo

rce

co

eff

icie

nt



Figure 3.58a: Axial force in the bottom chord at mid-span of the Boyne Viaduct


-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0


De

flec

tio

n c

oeff

icie

nt



Figure 3.56b: Vertical displacement at mid-span of the Boyne Viaduct subjected to

train Type 8 traversing at 30.3 m/s (109.1 km/hr)

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0


Ax

ial

forc

e c

oeff

icie

nt



Figure 3.57b: Axial force in the top chord at mid-span of the Boyne Viaduct

subjected to train Type8 traversing at 30.3 m/s (109.1 km/hr)

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0


Axia

l fo

rce

co

eff

icie

nt



Figure 3.58b: Axial force in the bottom chord at mid-span of the Boyne Viaduct

subjected to train Type8 traversing at 30.3 m/s (109.1 km/hr)


113

It is worth examining the Boyne Viaduct subjected to these two train models at

different speeds, in particular, the critical speed at which resonance of the bridge is

likely to occur due to repetitive vehicle loading. Examining Figure 3.51, the repetitive

distance Dw between railway carriages of train Type 1 is 20.3 m, while the repetitive

distance Dw between railway carriages of train Type 8 is 9.7 m. Using Equation

(3.46b) and assuming n = 1, the speeds of the two trains are:

( )1 3.123 20.3 63.4 m/s or 228.24 km/hr train Type 1wc f D= = × = (3.50a)

( )1 3.123 9.7 30.3 m/s or 109.08 km/hr train Type 8wc f D= = × = (3.50b)

The vertical displacement, axial force in the top chord and the axial force in the

bottom chord at mid-span of the Boyne Viaduct subjected to the two different train

types travelling at the speed computed in Equation (3.50) are shown in Figures 3.56 to

3.58, respectively. Inspecting the results in Figure 3.56 to 3.58, it can be seen that, on

this occasion, the dynamic coefficient for the Boyne Viaduct tends to undergo much

larger periodic oscillations compared with the results in Figure 3.53 to 3.55. It also

shows that the Boyne Viaduct is susceptible resonance generated by repetitive loaded

vehicles travelling at realistic speeds.

3.4 Discussion of results and Conclusion

To summarize, this chapter models a vehicle as a moving force i.e. only the

gravitational effects of the vehicle are taken into account. This model type is more

associated with large spanning bridges where the mass of the vehicle is often

substantially less than the mass of the bridge; thus, the inertia effects of the vehicle

are negligible (Fryba, 1996). In addition, if one is primarily interested in the dynamic

response of the bridge, then the moving force is a more suitable model to consider as

the time of execution is often less than that of other systems.


114

It was discovered in the preliminary stage of this thesis that the ANSYS finite element

program had limitations, such as not being able to apply a point force between the

nodes of an element; therefore, the author conceived the idea of applying nodal forces

(simple model) or nodal forces and nodal moments (exact model) as functions of time

to all the nodes along the bridge to simulate the moving constant force.

In Sections 3.2.2.1 to 3.2.2.2, it has been shown that the simple model is comparable

to the exact model provided that a suitable number of beam elements are used to

discretize the beam because Section 3.2.2.3 highlights the limitations of the simple

model as well as the ANSYS contact element method. The simple model and ANSYS

contact element model tend to lose accuracy as the number of elements in the model is

reduced. The beam deflection of the exact model is unaffected by the reduction in the

number of beam elements. However, the bending moment is affected by the number

of elements. Applying joint equilibrium at a particular node, one found that the

summation of the two internal moments at that node was equal to the applied moment

PG1 or PG2, depending on which element the moving force was located on. Section

3.2.2.4 then reveals that the internal forces from Equation (3.6) and (3.5) for the

simple and exact models are exactly the same as ANSYS produces for a pressure

applied between two consecutive nodes. The ANSYS internal forces are noticeably

different when the point force is applied at the nodes because ANSYS tends to neglect

the pressure loads when it computes the beam’s internal forces; hence, the shear force

and bending moments are stepped. Nevertheless, by applying joint equilibrium to a

node, one can say that the ANSYS solution is numerically correct.


115

Next using the developed methods, the author analyses a two-dimensional Boyne

Bridge subjected to single or multiple loads traversing at a constant speed. It is

observed that for a single moving force the maximum dynamic amplification factor,

DAF, occurs when the speed ratio α is 0.5, while for multiple moving forces the

maximum DAF occurred at α = 1.0. Despite these differences, both models tend to

show similarities at lower speeds, that is, at speeds ranging between 0.15 ≤ α ≤ 0.20

when a DAF value of 1.1 or greater is observed. It is shown that the results of the

three-dimensional Boyne Bridge are similar to results of the two-dimensional model

at low to medium speeds, which is, α < 0.4, and while at higher speeds the results

tend to diverge by as much as 10%. Like the two-dimensional model, the three-

dimensional model has a minimal DAF at approximately 0.27α ≈ .

It has been shown in Section 3.2.2 and 3.3.3 that resonance of the bridge can occur if

a critical speed is reached. This critical speed can be related to: (1) the traffic speed

using Equation (3.30); and (2) the repetitive vehicle loading of the bridge using

Equation (3.46b). In this thesis, the critical speed of the three-dimensional Boyne

Viaduct is 1814 km/hr using Equation (3.30), which is an unrealistic speed, while

using Equation (3.46b) and a repetitive distance of Dw of 5 m, the three-dimensional

Boyne Viaduct could be susceptible to resonance at 56.2 km/hr.

For the twin-track structures when the two trains have significantly different speeds

i.e. one train has a high speed while the other has a low speed, the DAFU can have a

value less than 1.0. As a final assessment, the three-dimensional Boyne Viaduct is

subjected to two different Eurocode (EN 1991-2, 2003) train models, in particular,

Type 1 travelling at 200 km/hr and Type 8 travelling at 100 km/hr. At these particular


116

speeds, one observes that the dynamic coefficient for both sets of results has a value of

1.0 or less on most occasions. However, when the speeds of both trains were changed

to the values computed in Equation (3.46), it was found that the Boyne Viaduct could

be susceptible to resonance generated by repetitive loaded vehicles travelling at

realistic speeds.

As a final note, the moving force system is not able to incorporate the inertia effects

of the vehicle, but frequently the inertia isn’t important. In addition, one can acquire

reasonable results on the dynamic behaviour of the bridge with minimum effort,

which is excellent in the early stages of the analysis.

Chapter 4 – Sprung mass represented by time varying stiffness matrices

117

Chapter 4

Sprung mass represented by time varying stiffness

matrices

4.1 Introduction

The primary aim of this chapter is to simulate the dynamic vertical response of a

vehicle traversing rigid rails and a railway bridge. This is achieved by using the

author’s wheel-rail contact element (WRC) to model the dynamic interaction between

a sprung wheel and the rail, using a Hertzian spring. The objective in creating this

element is to model the rail and wheel irregularities. The contact elements within the

ANSYS finite element program cannot model these irregularities. The wheel-rail

contact element incorporating irregularities is discussed in Chapter 6.

Many researchers such as Cheng et al. (1999), Yau et al. (1999) and Yang and Wu

(2001) have developed their own vehicle-bridge elements, where there is an unsprung

mass for the wheel and sprung mass for the vehicle body. In these systems, the wheel

is assumed to be in direct contact with the rail at all times; hence, the wheel and rail

have the same deflection and wheel-rail separation is not possible on a rigid rail. In

the author’s model, the wheel is represented by a sprung mass; thus, there is a


118

Hertzian spring between the wheel and the rail, which is simulated by means of the

WRC elements. In this case, the wheel and beam no longer have the same deflection.

The author’s technique involves modelling each wheel as a point mass and with the

Hertzian spring perpendicular to the surface of the rail. Each WRC element consists

of three stiffness matrices to simulate the action of the wheel on the flexible rail, but

only one stiffness matrix is required to represent the action of the wheel on the rigid

rail. The entries into the stiffness matrices depend on the position of a wheel on a

particular element and the element shape functions. The WRC elements use the

compression in the Hertzian spring at each time-step to calculate the contact force that

exists between the wheel and the rail. Wheel-rail separation occurs when compression

changes to positive extension in the spring, thus all stiffness matrices related to that

particular wheel are made equal to zero (Bowe & Mullarkey, 2005).

In addition to a vertical spring element, the author also develops longitudinal and

lateral spring elements that can take into account the braking and accelerating effect

of railway carriages as well as providing lateral support to the wheels of the carriage.

The braking forces are modelled using longitudinal spring elements located between

the wheels and flexible rail so that when the brakes are applied to the wheels, the

bridge feels the horizontal effects as the train decelerates.

This chapter also exposes the limitations of the ANSYS contact elements. Firstly, the

ANSYS contact elements have friction capabilities, but this cannot be used to model

the braking or accelerating effects of a wheel. Secondly, in three-dimensional models,

one must use a solid brick element to represent the rails, because the three-


119

dimensional ANSYS contact elements require a surface area. Thirdly, the three-

dimensional ANSYS contact elements neglect the lateral support of the wheels.

However, with the adaptation of the author’s lateral spring element between the

wheels and flexible rails, the wheels can be provided with adequate lateral support.

Furthermore, with the lateral spring element, the author can also examine lateral rail

irregularities, which are conducted in Chapter 6.

It should also be noted that any vehicle consisting of sprung masses must undergo a

static analysis prior to the transient analysis to ensure that the vehicle has settled

under its own weight at time t = 0 sec. Omitting this static analysis will cause sprung

masses in the model to experience free vibration as the transient analysis is initiated.

Following the validation of the WRC element by means of the ANSYS contact

element, the author studies a single sprung wheel as well as multiple railway carriages

traversing the Boyne Viaduct and compares the results of the bridge with the results

of the bridge for the moving loads examined in Chapter 3.


120

4.2 Development of the Wheel-Rail Contact Elements

4.2.1 Vertical Spring Element using time varying matrices

The author’s wheel-rail contact (WRC) element involves modelling a wheel as a mass

and a Hertzian spring with one node at the centre of the wheel and the other node on

the rail as illustrated in Figure 4.1. It is assumed that the Hertzian spring is always

perpendicular to the surface. Figure 4.1 shows local node 1 of the spring is in contact

with the rail and local node 2 of the spring is attached to the point mass.

Figure 4.1: Wheel modelled as a sprung mass

In Figure 4.2, the reader can see the sprung mass between nodes i-1 to i, as well as the

free-body diagram for the WRC element and one beam finite element representing

part of the rail. In ANSYS, the user can create a stiffness matrix for any pair of nodes

using the 12x12 MATRIX27 element. The author is obliged to use this facility to

input three additional stiffness matrices to simulate the Herztian spring. To simulate

the spring, three additional stiffness matrices are created between local node 1 and

local node 2 of the beam, between local node 1 of the beam and local node 2 of the

spring and finally between local node 2 of the beam and local node 2 of the spring.

2

1

⇒ Sprung mass

Wheel Centre

Rail

Wheel Centre

(a)

L

2x 3x 1ix − ix 1ix +1 0x =

2 31 x

y

x ct=

i 1i +1i −


121

Figure 4.2: (a) Sprung mass located between node i-1 to node i (b) free-body

diagram of the wheel-rails contact element

Figure 4.2 indicates the beam element coordinate system, where χ is positive along

the beam element, y is positive upward and z is positive outwards. The x-coordinate

is for the entire beam with the origin at the left hand support. The origin of the χ

coordinate system is at local node 1 of the beam. For two-dimensional problems, the

deflection in the x, y plane and rotation about z-axis are denoted as u, v and z

θ ,

respectively, while the subscripts L denote Hertzian spring. The beam element

coordinates will be used for this development. Recalling Equation (3.7), the

relationship between x and χ is as follows:

1ix xχ −= + (4.0)

The interaction between node 1 of the spring and the point of the beam element it

contacts is in accordance with Newton’s third law. The beam acts with a vertical force

of 1LyF on the spring; the spring acts with an equal and opposite force of 1Ly

F− on the

beam. From the free-body in Figure 4.2, the nodal forces in the wheel-rail contact

element can be expressed in the beam element coordinate system as follows:

(b)

ULx2

ULy2

1

1

1

2

2

1

1st stiffness matrix

2nd

stiffness matrix 3rd

stiffness matrix Hertzian

Spring

0

Mw

kH

y

x

z

1u 2u

2v1v

χ

i1-i2

ˆzθ

1ˆzθ

x ct=


122

1 1

1 12 11 2 12

2 22 21 2 22

2 2

Lx Lx

Ly LyL L

L L

Lx LxL L

Ly Ly

U F

U F

U F

U F

=

K KK U =

K K (4.1)

where only the x and y coordinates of the spring are taken into account for the axial

element; therefore 2LijK (i, j =1, 2) denotes a 2x2 stiffness matrix, LxiU , LyiU (i, j =1,

2) are the nodal displacement vector components and LxiF , LyiF (i, j =1, 2) are the

nodal force vector components of the Hertzian spring (also referred to as a link).

2 11LK is a symmetrical matrix, 2 22L

K is equal to 2 11LK , 2 12L

K is equal to 2 21LK , the

latter being equal to minus one times 2 11LK . From Figure 4.2 and Figure B.16a, one

can see that the Hertzian spring makes an angle o0α = (rotation about the y-axis) and

o90β = (rotation about the z-axis); therefore using Equation (B.97) the stiffness

matrix for the Hertzian spring is as follows (the third row and column that relate to the

z-axis of the axial element are omitted for this two-dimensional development):

2 11

0 0

0 1L Hk

=

K , is non-linear Hertzian spring stiffnessHk (4.2)

The vertical displacement at any point along the beam can be calculated as follows:

( ) ( ) ( ) ( ) ( )1 1 1 1 2 2 2 2ˆ ˆ

B z zv v N G v N Gχ χ θ χ χ θ χ= + + + (4.3)

where ( )1N χ , ( )1G χ , ( )2N χ and ( )2G χ are shape functions that are defined by

Equation (B.8) and plotted in Figure B.3 in Appendix B. The beam element

coordinate is defined by the symbol χ .


123

The length of the beam element is denoted by l. Since local node 1 of the spring is

located at position x ct= in the beam coordinate system, then its location in the beam

element coordinate system is equal to 1 1i ix x ct xχ − −= − = − ; hence:

( ) ( )1 1Ly B B iU v v ct xχ −= = − (4.4)

Equation (4.4) is only valid when the sprung mass is on the beam. Since it is assumed

that to the left of the left hand support and to the right of the right hand support are

rigid rails; hence, 1LyU becomes zero when the sprung mass is not on the beam. The

substitution of Equation (4.4) into Equation (4.1) gives the following:

( )2 1

1 12 11 2 12

2 22 21 2 22

2 2

Lx Lx

B i LyL L

Lx LxL L

Ly Ly

U F

v ct x F

U F

U F

−

−

=

K K

K K (4.5)

Since it is assumed that the Hertzian spring element remains perpendicular to the

surface at all times, the horizontal displacement on both nodes are equal

( )1 2 .Lx LxU U= In order to represent the vertical spring, the author returns to the exact

model for the moving load developed in Chapter 3. One rewrites Equations (3.16) and

(3.5) in Chapter 3 as:

( )( )( )( )

( )( )( )( )

1 1

2 21 1

2 2

2 20 0

2 2

l l

N N

G Gv vEI d m d

N N t

G G

χ χ

χ χχ χ

χ χχ

χ χ

′′ ′′ ∂ ∂

+ ′′ ∂ ∂

′′

∫ ∫

1

1

2

2

ˆ

ˆ

ˆ

ˆ

Q

M

Q

M

=

( )( )( )( )

( )

1

1

20

2

,

l

N

Gp x t d

N

G

χ

χχ

χ

χ

+

∫ (4.6a)

where


124

( )( )( )( )

( )

1

1

20

2

,

l

N

Gp x t d

N

G

χ

χχ

χ

χ

∫

( )( )( )( )

( )( )

1

1

1

20

2

l

i

N

Gct x Pd

N

G

χ

χδ χ χ

χ

χ

−

= − −

∫

( )( )( )( )

( )( )( )( )

1 1 1 1

1 1 1 1

1

2 1 2 1

2 1 2 1

i i

i i

Ly

i i

i i

N ct x N ct x

G ct x G ct xP F

N ct x N ct x

G ct x G ct x

− −

− −

− −

− −

− −

− − = =−

− − − −

(4.6b)

Equation (4.6b) is only valid when time t lies between 1ix

c

− and .ix

c The force

imparted to the beam by the spring is equal to 1LyF− located at position 1ict xχ −= − .

Expanding Equation (4.5), with the aid of Equation (4.2), for 1LyF then gives:

[ ] [ ] ( ) [ ] [ ]1 2 11 2 2 11 1 2 12 2 2 12 221 22 12 22Ly L Lx L B i L Lx L LyF U v ct x U U−= ⋅ + ⋅ − + ⋅ + ⋅K K K K

[ ] ( ) [ ]1 2H B i H Lyk v ct x k U−= ⋅ − + − ⋅

[ ]( )1

2

B i

H H

Ly

v ct xk k

U

−− = −

(4.7a)

By bringing the right hand side of Equation (4.6b) over to the left hand side, 1LyF

becomes positive and Equation (4.7a) can be rewritten as follows:

( )( )( )( )

( )( )( )( )

( )( )( )( )

[ ]( )

1 1 1 1 1 1

11 1 1 1 1 1

1 1

22 1 2 1 2 1

2 1 2 1 2 1

i i i

B ii i i

Ly Ly H H

Lyi i i

i i i

N ct x N ct x N ct x

v ct xG ct x G ct x G ct xF F k k

UN ct x N ct x N ct x

G ct x G ct x G ct x

− − −

−− − −

− − −

− − −

− − −

− − − − = = −

− − − − − −

(4.7b)

In Equation (4.7b), ( )1B iv ct x −− is replaced by the right hand side of Equation (4.3),

resulting in the following equation:

[ ]

1 1

1 1 1 1 1 1 2 2 2 2

1

2 2 2

2 2

ˆ ˆz z

Ly H H

Ly

N N

G G v N G v N GF k k

N N U

G G

θ θ

+ + +

= −

(4.8)


125

For convenience, in Equation (4.8) and in future equations ( )1 1iN ct x −− is replaced

by 1N , ( )1 1iG ct x −− is replaced by 1G etc. Equation (4.8) can also be rewritten as:

[ ]

1

1 1

1

1 1 1 1 2 2

21

2 2

22 2

2

ˆ0

0 0 0 0 1ˆ

z

Ly H H

z

Ly

vN N

G G N G N GvF k k

N N

G GU

θ

θ

= −

(4.9)

Expanding Equation (4.9) gives

[ ]

1

1 1 1 1 1 1 2 1 2 1

1

1 1 1 1 1 1 2 1 2 1

21

2 2 1 2 1 2 2 2 2 2

22 2 1 2 1 2 2 2 2 2

2

ˆ

ˆ

z

Ly H

z

Ly

vN N N N G N N N G N

G G N G G G N G G GvF k

N N N N G N N N G N

G G N G G G N G G GU

θ

θ

−

− = − −

(4.10)

Next, the author separates the right hand side of Equation (4.10) into three stiffness

matrices. The first matrix represents the stiffness matrix for local nodes 1 and 2 of the

beam element, the second matrix represents the stiffness matrix for local node 1 of the

beam and local node 2 of the spring element and the third matrix represents the

stiffness matrix for local node 2 of the beam and local node 2 of the spring element.

To facilitate the addition of the longitudinal spring element, which is developed in

Section 4.2.2, the author adds extra zero rows and zero columns to Equation (4.11).

[ ]

1

11 1 1 1 1 1 2 1 2

1 1 1 1 1 1 2 1 2 1

1

2

2 2 1 2 1 2 2 2 2 2

2 2 1 2 1 2 2 2 2 2

0 0 0 0 0 0 0

0 0

ˆ0 0

0 0 0 0 0 0 0

0 0

ˆ0 0

z

Ly H

z

u

vN N N N G N N N G

G G N G G G N G GF k

u

N N N N G N N N G v

G G N G G G N G G

θ

θ

=


126

[ ] [ ]

1 2

1 21 2

1 21 2

2 2

2 2

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

ˆ ˆ0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

z zH H

Lx Lx

Ly Ly

u u

v vN N

G Gk k

U U

U U

θ θ

+ − + −

− −

(4.11)

Next, one examines the equation of motion of the point mass, which is attached to

local node 2 of the spring element. This point mass has six degrees of freedom;

however, only the longitudinal, vertical and lateral motions are taken into account at

this time; therefore the equation of motion can be written as follows:

2

2 2

2

0 0 0

0 0

0 0 0

w Lx

w Ly Ly

w Lz

U

U F

U

= −

M

M

M

&&

&&

&&

(4.12a)

Hence, the force imparted to the point mass, representing the wheel, by the spring is

equal to 2LyF− . By bringing the right hand side of Equation (4.12a) to the left hand

side, 2LyF becomes positive and is derived by expanding Equation (4.5) as follows:

[ ] [ ] ( ) [ ] [ ]2 2 21 2 2 21 1 2 22 2 2 22 221 22 12 22Ly L Lx L B i L Lx L LyF U v ct x U U−= ⋅ + ⋅ − + ⋅ + ⋅K K K K

[ ] ( ) [ ]1 2H B i H Lyk v ct x k U−= − ⋅ − + ⋅

[ ]( )1

2

B i

H H

Ly

v ct xk k

U

−− = −

(4.12b)

Substituting Equation (4.3) into (4.12b) gives:


127

[ ]

1

1

1 1 2 2

22

2

2

ˆ0

0 0 0 0 1ˆ

z

Ly H H

z

Ly

v

N G N GvF k k

U

θ

θ

= − =

[ ][ ] [ ]

1

1

2 1 1 2 2 2

2

2

0 0

0 0

ˆ0 0

0 0

ˆ1 1

z

Ly H H Ly

z

v

F k N G N G k Uv

θ

θ

= − +

− −

=

[ ] [ ]

1 2

1 2

1 2

2 2

1 1 2 22 2

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

ˆ ˆ0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

z zH H

Lx Lx

Ly Ly

u u

v v

k kU U

N G N GU U

θ θ

= − + −

− −

[ ]

1

1

1

2

2

0 0 0 0 0 0

0 0 0 0 0 0

ˆ0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 1 0

0 0 0 0 0 0

zH

Lx

Ly

u

v

kU

U

θ

+

−

(4.12c)

The right hand side of Equation (4.12c) contains three matrix terms, and the third term

by itself represents the effects of the rigid rails. For programming purposes Equations

(4.11) and (4.12c) are combined, resulting in symmetrical stiffness matrices for the

Hertzian spring, updated at each time-step.

The wheel-rail contact force in the spring is evaluated by multiplying the extension by

the spring stiffness. The Hertzian extension is given by Equation (B.109) where α is

equal to 0o

and β is equal to 90o

for the vertical spring and Equation (4.4) has also

been incorporated, giving:


128

[ ] [ ]1 2

1 2

extension cos cos sin cos cos sinLx Lx

Ly Ly

U U

U Uβ α β β α β

= − +

(4.13a)

[ ]( )

1

1

2

2

extension 0 1 0 1

Lx

B i

Lx

Ly

U

v ct x

U

U

−

−

= −

(4.13b)

In the model, the extension in the spring is calculated at each time-step to determine if

the wheel is in contact with the rail. A negative extension (compression) indicates that

contact exists between the wheel and the rail, while a positive extension (tension)

means that there is no contact; thus all stiffness matrices related to that particular

wheel are set equal to zero when the extension is positive.

4.2.2 Longitudinal Spring Element using time varying matrices

Using a similar methodology to that of the previous section, the author now develops

a longitudinal spring element between wheel centre and the rail surface to simulate

both the accelerating and braking effects of the wheel. This spring element is

tangential to the vertical spring element and is parallel with the beam surface as

shown in Figure 4.3. This particular beam element is also between node i-1 to node i,

with the entire beam shown in Figure 4.2a.

Figure 4.3: Free-body diagram of the longitudinal spring element

1

2

2u

2v

i

1

1

0

y

x

z

1u

1v

χ

1-i

ULx2

ULy2

2

1

Longitudinal

Spring Mw

kH

2ˆz

θ

1ˆz

θ

x ct=


129

Recalling Equation (4.1), the nodal forces in the longitudinal spring element can be

expressed in the beam element coordinate system as follows:

1 1

1 12 11 2 12

2 22 21 2 22

2 2

Lx Lx

Ly LyL L

L L

Lx LxL L

Ly Ly

U F

U F

U F

U F

=

K KK U =

K K (4.14)

where 2LijK (i, j =1, 2) denotes a 2x2 stiffness matrix. From Figure B.16b, the

longitudinal spring makes an angle o0α = (rotation about the y-axis) and o0β =

(rotation about the z-axis), therefore using Equation (B.98) the stiffness matrix for the

longitudinal spring becomes (again the third row and column that relate to the z-axis

of the axial element are omitted):

2 11

1 0

0 0L H

k

=

K (4.15)

The longitudinal displacement at any point along the beam is calculated as follows:

( ) ( ) ( )1 1 2 2Bu u H u Hχ χ χ= + (4.16)

where ( )1H χ and ( )2H χ are shape functions that are defined by Equation (B.47)

and plotted in Figure B.7 in Appendix B. The beam element coordinate is again

defined by the symbol χ . Since local node 1 of the longitudinal spring is located at

position 1ict xχ −= − on the beam, one can state the following:


130

( )1 1Lx B iU u ct x −= − (4.17)

Equation (4.17) is only valid when the sprung mass is on the beam; hence, when the

sprung mass is on rigid rails and not on the beam, 1LxU becomes zero. The

substitution of Equation (4.17) into Equation (4.14) gives the following:

( ) 11

122 11 2 12

222 21 2 22

22

LxB i

LyLyL L

LxLxL L

LyLy

Fu ct x

FU

FU

FU

−−

=

K K

K K (4.18)

Since it is assumed that the longitudinal spring remains parallel to the surface at all

times, the vertical displacements on both nodes are equal. On this occasion, to

represent the longitudinal motion of the beam, one requires Equation (B.53) and (3.9), giving:

( )( )

( )( )

( )( )

( )2

1 1 11

2

2 2 20 0 02

ˆ,

ˆ

l l lH H HPu uEA d m d p x t d

H H Hx t P

χ χ χχ χ χ

χ χ χ

′ ∂ ∂ + = +

′ ∂ ∂ ∫ ∫ ∫ (4.19a)

where

( )( )

( )1

20

,

l Hp x t d

H

χχ

χ

∫

( )( )

( )( )1

1

20

l

i

Hct x Pd

H

χδ χ χ

χ −

= − − ∫

( )( )

( )( )

1 1 1 1

1

2 1 2 1

i i

Lx

i i

H ct x H ct xP F

H ct x H ct x

− −

− −

− − = =−

− −

(4.19b)

Equation (4.19) is only valid when time t lies between 1ix

c

− and .ix

c The force

imparted to the beam by the longitudinal spring is equal to 1LxF− located at

position 1ict xχ −= − . Expanding Equation (4.18), using Equation (4.15), for 1Lx

F gives:


131

[ ] ( ) [ ] [ ] [ ]1 2 11 1 2 11 2 2 12 2 2 12 211 12 11 12Lx L B i L Ly L Lx L LyF u ct x U U U−= ⋅ − + ⋅ + ⋅ + ⋅K K K K

[ ] ( ) [ ]1 2H B i H Lxk u ct x k U−= ⋅ − + − ⋅

[ ]( )1

2

B i

H H

Lx

u ct xk k

U

− − = −

(4.20a)

By bringing the right hand side of Equation (4.19b) over to the left hand side, 1LxF


( )( )

( )( )

( )( )

[ ]( )1 1 1 1 1 1 1

1 1

2 1 2 1 2 1 2

i i i B i

Lx Lx H H

i i i Lx

H ct x H ct x H ct x u ct xF F k k

H ct x H ct x H ct x U

− − − −

− − −

− − − − = = −

− − − (4.20b)

In Equation (4.20b), ( )1B iu ct x −− can be replaced by the right hand side of Equation

(4.16), resulting in the following equation:

[ ] 1 1 2 21 1

1

22 2

Lx H H

Lx

u H u HH HF k k

UH H

+ = −

(4.21)

For convenience again, in Equation (4.21) ( )1 1iH ct x −− has been replaced by 1H and

( )2 1iH ct x −− has been replaced by 2H . Equation (4.21) can also be rewritten as:

[ ]1

1 1 1 2

1 2

2 2

2

0

0 0 1Lx H H

Lx

uH H H H

F k k uH H

U

= −

(4.22)


[ ]1

1 1 1 1 2 1

1 2

2 2 1 2 2 2

2

Lx H

Lx

uH H H H H H

F k uH H H H H H

U

−

= −

(4.23)


132

As before, the author separates the right hand side of Equation (4.23) into three

stiffness matrices. The first matrix represents the stiffness matrix for local nodes 1 and

2 of the beam element, the second matrix represents the stiffness matrix for local node

1 of the beam and local node 2 of the longitudinal spring element and the third matrix

represents the stiffness matrix for local node 2 of the beam and local node 2 of the

longitudinal spring element.

[ ]

11 1 1 1 2

1

1

1

22 2 1 2 2

2

2

0 0 0 0

0 0 0 0 0 0 0

ˆ0 0 0 0 0 0 0

0 0 0 0

0 0 0 0 0 0 0

ˆ0 0 0 0 0 0 0

z

Lx H

z

uH H H H H

v

F kuH H H H H

v

θ

θ

=

[ ] [ ]

1 21 1

1 2

1 2

2 2

2 2

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

ˆ ˆ0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

z zH H

Lx Lx

Ly Ly

u uH H

v v

k kU U

U U

θ θ

+ − + −

− −

(4.24)

Similar to Section 4.2.1, one examines the equation of motion of the point mass,

which is attached to local node 2 of the longitudinal spring element and can be written

as follows:

2 2

2

2

0 0

0 0 0

0 0 0

w Lx Lx

w Ly

w Lz

U F

U

U

− =

M

M

M

&&

&&

&&

(4.25a)

The force imparted to the point mass by the longitudinal spring is equal to 2LxF− . By

bringing the right hand side of Equation (4.25a) to the left hand side, 2LxF becomes

positive and is derived by expanding Equation (4.18) as follows:


133

[ ] ( ) [ ] [ ] [ ]2 2 21 1 2 21 2 2 22 2 2 22 211 12 11 12Lx L B i L Ly L Lx L LyF u ct x U U U−= ⋅ − + ⋅ + ⋅ + ⋅K K K K

[ ] ( ) [ ]1 2H B i H Lxk u ct x k U−= − ⋅ − + ⋅

[ ]( )1

2

B i

H H

Lx

u ct xk k

U

− − = −

(4.25b)


[ ]1

1 2

2 2

2

0

0 0 1Lx H H

Lx

uH H

F k k u

U

= − =

[ ][ ] [ ]1

2 1 2 2

2

0 0

0 0

0 0

1 1

0 0

Lx H H Lx

uF k H H k U

u

= − + =

− −

[ ] [ ]

1 2

1 2

1 2

1 12 2

2 2

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

ˆ ˆ0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

z zH H

Lx Lx

Ly Ly

u u

v v

k kH HU U

U U

θ θ

= − + −

− −

[ ]

1

1

1

2

2

0 0 0 0 0 0

0 0 0 0 0 0

ˆ0 0 0 0 0 0

0 0 0 1 0 0

0 0 0 0 0 0

0 0 0 0 0 0

zH

Lx

Ly

u

v

kU

U

θ

+

−

(4.25c)

For programming purposes Equations (4.24) and (4.25c) are combined, resulting in

symmetrical stiffness matrices for the longitudinal spring element, updated at each

time-step.


134

Similar to Section 4.2.1, the braking force or horizontal axial force in the spring

element is evaluated using the extension in the spring element. On this occasion, one

is only concerned with the horizontal nodal displacements; thus α and β are both

equal to 0o. The braking force felt by the wheel is given by Equation (B.109), which

incorporates Equation (4.17) as follows:

[ ] [ ]1 2

1 2


Ly Ly

U U


= − +

(4.26a)

[ ]

( )1

1

2

2

extension 1 0 1 0

B i

Lx

Lx

Ly

u ct x

U

U

U

−−

= −

(4.26b)

According to Newton’s second law of motion, as the train accelerates, additional

horizontal forces must be applied to the train using Equation (4.27).

, , w w b b v v

F M a F M a F M a= = = (4.27)

where , , w b c

F F F are additional forces and , , w b v

M M M are the mass of the wheels,

bogies and vehicle body of the train, respectively, and a is the acceleration. In order

for the wheels of the train to experience these additional horizontal forces, horizontal

suspension springs are then required between the bogies and wheels and between the

bogies and vehicle body.


135

4.2.3 Lateral Spring Element using time varying matrices

This section develops the lateral spring element that provides lateral support to the

wheels of the vehicle in three-dimensional analyses. In addition, these elements allow

the author to examine lateral loading from the wind. In Figure 4.4a, the author shows

a free-body diagram for the lateral spring element and one beam element (between

node i-1 to node i of the beam shown in Figure 4.2a) representing part of the rail. As

before, to simulate the lateral spring one uses additional stiffness matrices between

local node 1 and local node 2 of the beam, between local node 1 of the beam and local

node 2 of the spring and finally between local node 2 of the beam and local node 2 of

the spring. In Figure 4.4a the coordinate system is such that χ positive along the

beam element, y positive upward and z positive outwards. The origin of the coordinate

system is at local node 1 of the beam. The deflection in the x, y and z plane and

rotation about the y and z plane are denoted as u, v, w, y

θ and z

θ , respectively.

Figure 4.4a: Free-body diagram of the lateral spring element (x-y plane)

Figure 4.4b shows the lateral spring element in the x-z plane. The development of the

lateral spring in this plane is made easier because this diagram is very similar to

Figure 4.2a. The main difference is a change of direction of the nodal moments. The

y

x

z

1

1

2

1

2

1

2w 2v

2u

1w2v

2u

2ˆ

yθ

1ˆ

yθ 1ˆz

θ2

ˆz

θ

2LzU

2LxU

χ

Mw i i -1

x ct=


136

updated coordinate system now has χ positive along the beam element, y positive

inwards and z positive upwards; hence, this lateral spring can also be represented as a

two-dimensional problem such that the deflection in the x, z plane and rotation about

y-axis are denoted as u, w and y

θ , respectively.

Figure 4.4b: Free-body diagram of the lateral spring element (x-z plane)

From the free-body in Figure 4.4b, the nodal forces in the lateral spring element can

be expressed in the beam element coordinate system as follows:

1 1

2 11 2 12 1 1

2 21 2 22 2 2

2 2

Lx Lx

L L Lz Lz

L L

L L Lx Lx

Lz Lz

U F

U F

U F

U F

=

K KK U =

K K (4.28)

where Lij

K (i, j =1, 2) denotes a 2x2 stiffness matrix. From Figure 4.4a and Figure

B.16b, one can see that the Hertzian spring makes an angle o270α = (rotation about

the y-axis) and o0β = (rotation about the z-axis); therefore using Equation (B.99) the

stiffness matrix for the lateral spring becomes (the second row and column that relate

to the y-axis of the axial element are omitted):

ULx2

ULz2

1

1

1

2

1

Lateral

Spring

0

Mw

kH

z

x

y

1u 2u

2w1w

χ

2

1st stiffness matrix

2nd

stiffness matrix 3rd

stiffness matrix

i -1 i

2ˆ

yθ1ˆ

yθ

x ct=


137

2 11

0 0

0 1L H

k

=

K (4.29)

The lateral displacement at any point along the beam can be calculated as follows:

( ) ( ) ( ) ( ) ( )1 1 1 1 2 2 2 2ˆ ˆ

B y yw w N G w N Gχ χ θ χ χ θ χ= − + − (4.30)

where ( )1N χ , ( )1G χ , ( )2N χ and ( )2G χ are shape functions that are defined by

Equation (B.8) and plotted in Figure B.3 in Appendix B. It should be noted that

Equation (4.30) corresponds to Equation (B.28), with its negative rotational shape

functions. As before, the beam element coordinate is defined by the symbol χ .

The length of the beam element is denoted by l and the distance travelled along the

element is given by 1ict xχ −= − , travelling from left to right. Since local node 1 of the

lateral spring is located at position x ct= on the beam, one can state the following:

( ) ( )1 1Lz B B iU w w ct xχ −= = − (4.31)

Equation (4.31) is only valid when the point mass, representing the wheel, is on the

beam. When the wheel is not on the beam, 1LzU becomes zero. The substitution of

Equation (4.31) into Equation (4.28) gives the following:

( )2 1

2 11 2 12 1 1

2 21 2 22 2 2

2 2

Lx Lx

L L B i Lz

L L Lx Lx

Lz Lz

U F

w ct x F

U F

U F

−

−

=

K K

K K (4.32)


138

Since it is assumed that the lateral spring element remains parallel to the z-axis at all

times, the horizontal displacement on both nodes are equal ( )1 2 .Lx LxU U= In order to

represent the lateral spring, the author rewrites Equation (B.34) in Appendix B, with

the right-hand side fully evaluated in Equation (B.37) as follows:

( )( )( )( )

( )( )( )( )

1 1

2 21 1

2 2

2 20 0

2 2

l l

N N

G Gw wEI d m d

N N t

G G

χ χ

χ χχ χ

χ χχ

χ χ

′′ ′′− −∂ ∂

+ ′′ ∂ ∂

′′− −

∫ ∫

1

1

2

2

ˆ

ˆ

ˆ

ˆ

Q

M

Q

M

=

( )( )( )( )

( )

1

1

20

2

,l

N

Gp x t d

N

G

χ

χχ

χ

χ

− + −

∫

(4.33a)

where, with the aid of Equation (3.16) is

( )( )( )( )

( )

1

1

20

2

,l

N

Gp x t d

N

G

χ

χχ

χ

χ

− −

∫

( )( )( )( )

( )

1

1

20

2

l

N

Gx ct Pd

N

G

χ

χδ χ

χ

χ

− = −

−

∫

( )( )( )( )

( )( )( )( )

1 1 1 1

1 1 1 1

1

2 1 2 1

2 1 2 1

i i

i i

Lz

i i

i i

N ct x N ct x

G ct x G ct xP F

N ct x N ct x

G ct x G ct x

− −

− −

− −

− −

− − − − − −

= =− − −

− − − −

(4.33b)

Equation (4.33b) is only valid when time t lies between 1ix

c

− and .ix

c The force

imparted to the beam by the lateral spring element is equal to 1LzF− located at

position 1ict xχ −= − . Expanding Equation (4.32), with the aid of Equation (4.29), for

1LzF then gives:

[ ] [ ] ( ) [ ] [ ]1 2 11 2 2 11 1 2 12 2 2 12 221 22 12 22Lz L Lx L B i L Lx L LzF U w ct x U U−= ⋅ + ⋅ − + ⋅ + ⋅K K K K

[ ] ( ) [ ]1 2H B i H Lzk w ct x k U−= ⋅ − + − ⋅

[ ]( )1

2

B i

H H

Lz

w ct xk k

U

−− = −

(4.34a)


139

By bringing the right hand side of Equation (4.34a) over to the left hand side, 1LzF


( )( )( )( )

( )( )( )( )

( )( )( )( )

[ ]( )

1 1 1 1 1 1

1 1 1 1 1 1 1

1 1

2 1 2 1 2 1 2

2 1 2 1 2 1

i i i

i i i B i

Lz Lz H H

i i i Lz

i i i

N ct x N ct x N ct x

G ct x G ct x G ct x w ct xF F k k

N ct x N ct x N ct x U

G ct x G ct x G ct x

− − −

− − − −

− − −

− − −

− − −

− − − − − − − = = −

− − − − − − − − −

(4.34b)

In Equation (4.34b), ( )1B iw ct x −− is replaced by the right hand side of Equation

(4.30), resulting in the following equation:

[ ]

1 1

1 1 1 1 1 1 2 2 2 21

2 2 2

2 2

ˆ ˆy y

Lz H H

Lz

N N

G G w N G w N GF k k

N N U

G G

θ θ

− − − + −

= −

− −

(4.35)

As before, in Equation (4.35) ( )1 1iN ct x −− has been replaced by 1N , ( )1 1iG ct x −− has

been replaced by 1G etc. Equation (4.35) can also be rewritten as:

[ ]

1

1 1

1

1 1 1 1 2 2

1 2

2 2

22 2

2

ˆ0

0 0 0 0 1ˆ

y

Lz H H

y

Lz

wN N

G G N G N GF k k w

N N

G GU

θ

θ

− − − −

= −

− −

(4.36)


[ ]

1

1 1 1 1 1 1 2 1 2 1

1

1 1 1 1 1 1 2 1 2 1

1 2

2 2 1 2 1 2 2 2 2 2

22 2 1 2 1 2 2 2 2 2

2

ˆ

ˆ

y

Ly H

y

Lz

wN N N N G N N N G N

G G N G G G N G G GF k w

N N N N G N N N G N

G G N G G G N G G GU

θ

θ

− − −

− − − = − − − − − −

(4.37)


140

The right hand side of Equation (4.37) is developed further, giving rise to the

following equation, which contain the three stiffness matrices used by the lateral

elements:

[ ]

1

11 1 1 1 1 1 2 1 2

11 1 1 1 1 1 2 1 2

1

2

22 2 1 2 1 2 2 2 2

2 2 1 2 1 2 2 2 2 2

0 0 0 0 0 0 0

0 0

ˆ0 0

0 0 0 0 0 0 0

0 0

ˆ0 0

y

Lz H

y

u

wN N N N G N N N G

G G N G G G N G GF k

w

vN N N N G N N N G

G G N G G G N G G

θ

θ

− − − − −

= − −

− − −

[ ] [ ]

1 2

1 21 2

1 21 2

2 2

2 2

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0

ˆ ˆ0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

y yH H

Lx Lx

Lz Lz

u u

w wN N

G Gk k

U U

U U

θ θ

− −

+ − + − − −

(4.38)

There are three matrix terms on the right hand side of Equation (4.38). The first term

represents the stiffness matrix along the beam element (between local node 1 and

local node 2 of the beam), the second term represents the stiffness matrix for local

node 1 of the beam and local node 2 of the lateral spring element and the third term

represents the stiffness matrix for local node 2 of the beam and local node 2 of the

lateral spring element.

Next, the equation of motion of the point mass is examined, which is attached to local

node 2 of the lateral spring element. The point mass has six degrees of freedom;

however, only the longitudinal, vertical and lateral motions are taken into account at

this time; therefore the equation of motion can be written as follows:


141

2

2

2 2

0 0 0

0 0 0

0 0

w Lx

w Ly

w Lz Lz

U

U

U F

= −

M

M

M

&&

&&

&&

(4.39a)

Hence, the force imparted to the point mass, representing the wheel, by the spring is

equal to 2LzF− . By bringing the right hand side of Equation (4.39a) to the left hand

side, 2LzF then becomes positive and is derived by expanding Equation (4.32) as follows:

[ ] [ ] ( ) [ ] [ ]2 2 21 2 2 21 1 2 22 2 2 22 221 22 12 22Lz L Lx L B i L Lx L LzF U w ct x U U−= ⋅ + ⋅ − + ⋅ + ⋅K K K K

[ ] ( ) [ ]1 2H B i H Lzk w ct x k U−= − ⋅ − + ⋅

[ ]( )1

2

B i

H H

Lz

w ct xk k

U

−− = −

(4.39b)


[ ]

1

1

1 1 2 2

2 2

2

2

ˆ0

0 0 0 0 1ˆ

y

Lz H H

y

Lz

w

N G N GF k k w

U

θ

θ

− − = − =

[ ][ ] [ ]

1

1

2 1 1 2 2 2

2

2

0 0

0 0

ˆ0 0

0 0

ˆ

1 1

y

Lz H H Lz

y

w

F k N G N G k Uw

θ

θ

= − − − + = − −

[ ] [ ]

1 2

1 2

1 2

2 2

1 1 2 22 2

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

ˆ ˆ0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

y yH H

Lx Lx

Lz Lz

u u

w w

k kU U

N G N GU U

θ θ

= − + − − −

− −


142

[ ]

1

1

1

2

2

0 0 0 0 0 0

0 0 0 0 0 0

ˆ0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 1

yH

Lx

Lz

u

w

kU

U

θ

+ −

(4.39c)

The right hand side of Equation (4.39c) contains three matrix terms, and the third term

by itself represents the effects of the rigid rails. For programming purposes Equations

(4.38) and (4.39c) are combined, resulting in symmetrical stiffness matrices for the

lateral spring element, updated at each time-step.

The wheel-rail contact force in the spring is evaluated by multiplying the extension by

the spring stiffness. The Hertzian extension is given by Equation (B.109) where α is

equal to 270o


for the lateral spring as well as incorporating

Equation (4.4), giving:

[ ] 1

1

extension cos cos cos sinLx

Lz

U

Uβ α β α

= − −

[ ] 2

2

cos cos cos sinLx

Lz

U

Uβ α β α

+ −

(4.40a)

[ ]( )

1

1

2

2

extension 0 1 0 1

Lx

B i

Lx

Lz

U

w ct x

U

U

−

−

= −

(4.40b)


143

In Figure 4.5, the author presents the computational scheme for the wheel-rail contact

element, which includes the vertical, longitudinal and lateral spring elements

developed in Sections 4.2.1 to 4.2.3.

Figure 4.5: Computational scheme of the wheel-rail contact element


144

4.2.4 Multiple wheels using the WRC element

Until now, one has only considered a single sprung wheel traversing the beam using

the wheel-rail contact element. With the aid of Equation (3.38) and substituting

Equation (4.0) into (4.6), the equation of motion for a single beam element between

nodes i-1 to i, subjected to all the wheels of the train can be written as follows:

( )( )( )( )

( )( )( )( )

1 1

2 21 1

2 2

2 20 0

2 2

l l

N N

G Gv vEI d m d

N N t

G G

χ χ

χ χχ χ

χ χχ

χ χ

′′ ′′ ∂ ∂

+ ′′ ∂ ∂

′′

∫ ∫

1

1

2

2

ˆ

ˆ

ˆ

ˆ

Q

M

Q

M

=

( )( )( )( )

( ) ( ) ( ) ( ) ( )

1

2 21

1 , , 1 = 1 = 120

2

1 1 1c

l N

i w w w m j kj k m

N

Gct x D j B k A m P d

N

G

χ

χδ χ χ

χ

χ

−=

− − − + − + − + −

∑ ∑ ∑∫ (4.41)

where Nc is the number of carriages, Aw is the distance between the two axles of a

single bogie, Bw is the distance between centres of two bogies of a single carriage and

Dw is the distance from the front wheel of each carriage, c is the speed of the train, t is

time and ( ), ,m j kP is the weight of the wheels.

Let’s now begin by examining the Dirac delta function in Equation (4.41), i.e. the

position the m-th wheel, of the k-th bogie, and the j-th carriage of the train is equal to:

( ) ( ) ( ) ( )1 1 1 1 0i w w wct x D j B k A mχ −− − + − + − + − = (4.42a)

( ) ( ) ( ) ( )1 1 1 1i w w wct x D j B k A mχ −= − − − − − − − (4.42b)


145

For multiple wheels acting on a single beam element, the second term on the right

hand side of Equation (4.41) can be written as follows:

( )( )( )( )

( ) ( ) ( ) ( ) ( )

1

2 21

1 , , 1 = 1 = 120

2

1 1 1c

l N

i w w w m j kj k m

N

Gct x D j B k A m P d

N

G

χ

χδ χ χ

χ

χ

−=

− − + − + − + − =

∑ ∑ ∑∫

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( )

1 1

2 21 1

1 , , 1 = 1 m = 1 2 1

2 1

1 1 1

1 1 1

1 1 1

1 1 1

c

i w w w

Ni w w w

Ly m j k

j k i w w w

i w w w

N ct x D j B k A m

G ct x D j B k A mF

N ct x D j B k A m

G ct x D j B k A m

−

−

= −

−

− − − − − − −

− − − − − − − −

− − − − − − − − − − − − − −

∑ ∑ ∑ (4.43)

It should be noted that Equation (4.43) is only valid when time t lies between

( ) ( ) ( )1 1 1 1i w w w

x D j B k A m

c

− + − + − + − and

( ) ( ) ( )1 1 1i w w w

x D j B k A m

c

+ − + − + −.

Using the same development from Equation (4.7) to (4.12c), the three additional

stiffness matrices for the wheel-rail contact element subjected to multiple wheels are

derived.


146

4.3 Validating the Wheel-Rail Contact Elements

In order to validate the system, the author compares results from the use of the WRC

element with results from the use of the node-to-surface contact element in ANSYS

(CONTAC48) as well as with results from the literature. In all cases, unless otherwise

stated, the gravitational and damping effects of the beam are ignored, time t is

arranged in such a manner that the wheel (front wheel in Section 4.3.8) is at the left

support at t = 0 sec, the initial displacement and velocity of the beam is equal to zero,

the beam is discretized into 10 beam elements and the Newmark-β time integration

scheme (Bathe, 1996) with 200 equal time steps is used to solve the transient analysis.

It is also important to note that the Hertzian spring and sprung masses in each model

must undergo a static analysis prior to the execution of the transient analysis,

otherwise the masses will go into free vibration as the transient analysis is initiated.

4.3.1 Wheel as a moving sprung load traversing a cantilever beam

The author simulates a moving sprung load traversing a cantilever beam from the

fixed to the free end, and vice-versa as shown in Figure 4.6. The beam has a length L

of 7.62 m, flexural rigidity EI of 9.47 x 106 Nm

2 and mass per unit length m of 46

kg/m. Using Equation (C.27) the first natural frequency of this particular beam is

1 27.49 rad/sec,w = while the moving sprung load P is - 25.79 kN and it traverses the

beam at a constant speed c of 50.8 m/s. These properties are also adopted by Akin &

Mofid (1989). For this particular example, the author uses 200 equal time steps to

solve the transient analysis. The Hertzian spring stiffness kH is 1.4 x 106 kN/m, which

is a reasonably stiff value (Esveld, 2001).


147

Figure 4.6 Moving sprung load traversing a cantilever beam: (a) fixed-free beam;

(b) free-fixed beam

The vertical displacements at the free-end of the fixed-free and free-fixed cantilever

beam subjected to a moving sprung load, traversing from left to right, are plotted in

Figures 4.7a and 4.7b, respectively. One can see that the numerical results for the

WRC element and the ANSYS contacts agree with each other and with Figure 3.12.

-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

WRC Element

ANSYS CONTAC48

-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

WRC Element

ANSYS CONTAC48

Figure 4.7: Vertical displacement at the free-end of a cantilever beam subjected to

a moving sprung load: (a) left hand side is fixed; (b) right hand side is fixed

(a)

(b)

L

c

L

c (a) (b)

P P

Hk

Hk


148

4.3.2 Wheel as a moving sprung mass traversing a cantilever beam

Using the same beam and vehicle properties as in Section 4.3.1, the author now

examines a moving sprung mass traversing a cantilever beam from the fixed to the

free end, and from the free to the fixed end. On this occasion, the sprung mass Mw is

2629 kg, such that is 7.5w

M ml and a Hertzian spring stiffness kH of 1.4 x 106 kN/m

is used.

The vertical displacement at the free-end of the fixed-free and free-fixed cantilever

beam subjected to a moving sprung mass can be seen in Figures 4.8a and 4.8b,

respectively. Again, the results from using the WRC element agree well with the

results from using the ANSYS contact element CONTAC48.

-0.08

-0.06

-0.04

-0.02

0.00

0.02

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

WRC Element

ANSYS CONTAC48

-0.40

-0.35

-0.30

-0.25

-0.20

-0.15

-0.10

-0.05

0.00

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

WRC Element

ANSYS CONTAC48

Figure 4.8: Vertical displacement at the free-end of a cantilever beam subjected to

a moving sprung mass: (a) left hand side is fixed; (b) right hand side is fixed

(a)

(b)


149

In addition to examining the beam deflection, the author also examines the contact

force between the moving sprung mass and the fixed-free and free-fixed cantilever

beam, which are presented in Figures 4.9a and 4.9b. It should be noted that a negative

value indicates a compressive force between the wheel and the rail, while a zero value

of force indicates wheel-rail separation and all stiffness matrices related to that wheel

in the WRC element are set equal to zero. Therefore, one can conclude from Figure

4.9a that the wheel tends to lose contact with the beam at approximately time t = 0.08

sec and re-establishes contact briefly again at time t = 0.111 sec and 0.135 sec for this

particular speed and Hertzian spring stiffness.

-2.0

-1.5

-1.0

-0.5

0.0

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Co

nta

ct

forc

e / S

tati

c w

eig

ht

WRC Element

ANSYS CONTAC48

-16

-14

-12

-10

-8

-6

-4

-2

0

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Co

nta

ct

forc

e / S

tati

c w

eig

ht

WRC Element

ANSYS CONTAC48

Figure 4.9: Contact force between the wheel and rail: (a) left hand side is fixed;

(b) right hand side is fixed

(a)

(b)


150

Examining Figure 4.9b, one can conclude that the wheel tends to lose contact with the

beam on several occasions at the start of the analysis and as the wheel moves beyond

mid-span of the beam it gradually increases its wheel-rail compressive force. At time t

= 0.13 sec, the wheel-rail contact force is 15 times larger than the static weight of the

wheel. At time t = 0.14 sec, wheel-rail separation occurs. With the knowledge of

wheel-rail separation occurring in both beam models, the author finally plots the

vertical displacement of the moving sprung mass as a function of time in Figures

4.10a and 4.10b, respectively. It can be seen in Figure 4.10b that the wheel

experiences a positive upward displacement at time t = 0.14 sec, which explains the

loss of contact in Figure 4.9b.

-0.08

-0.06

-0.04

-0.02

0.00

0.02

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

WRC Element

ANSYS CONTAC48

Beam free end deflection - Figure 4.8a

-0.40

-0.30

-0.20

-0.10

0.00

0.10

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

WRC Element

ANSYS CONTAC48

Beam free end deflection - Figure 4.8b

Figure 4.10: Vertical displacement of the sprung mass as it travels along a

cantilever beam: (a) left hand side is fixed; (b) right hand side is fixed

(a)

(b)


151

4.3.3 Wheel as a moving sprung load traversing a simply supported beam

The bridge and vehicle properties adopted in the following examples in Sections 4.3.3

to 4.3.7 are the same as those of Yang and Wu (2001). The bridge has a length L of

25m, Young’s modulus of elasticity E of 2.87x106 kN/m

2, moment of inertia I of 2.9

m4, mass per unit length m of 2.303 t/m and a Poisson’s ratio ν of 0.2. In addition, the

wheel traverses the bridge at a constant speed c of 27.78 m/s (100 km/hr), the

Hertzian spring stiffness kH is 1595 kN/m and 200 equal time steps are used to solve

the transient analysis.

Figure 4.11 presents a simply supported beam, which is subjected to a moving sprung

load P travelling at speed c from left to right. As in Section 4.3.1, the wheel is given a

zero mass and the point force P of 56.4 kN is attached to the centre of the wheel i.e.

local node 2 of the spring.

Figure 4.11: Simply supported beam subjected to a moving sprung load

The vertical displacement, acceleration and bending moment at mid-span of the beam

as a function of time can be seen in Figures 4.12a, 4.12b and 4.12c, respectively. In

this example, the author only examines the internal forces at local node 2 of the

element at mid-span of the beam; however, one is aware that this internal force differs

slightly with local node 1 of the other element at mid-span. The reader should refer

back to Section 3.2.2.2 to 3.2.2.4 for an explanation of the differences between LN1

L


152

and LN2. In each plot, the author compares the results obtained using the WRC

element with results from using the ANSYS contact element and it can be observed

that both systems give very similar results.

-0.0025

-0.0020

-0.0015

-0.0010

-0.0005

0.0000

0.0005

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

WRC Element

ANSYS CONTAC48

-0.4

-0.2

0.0

0.2

0.4

0.6

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al A

cc

ele

rati

on

(m

/s2)

WRC Element ANSYS CONTAC48

0

100

200

300

400

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Be

nd

ing

Mo

me

nt

(kN

m)

WRC Element (Element5_LN2)

ANSYS CONTAC48 (Element5_LN2)


moment (local node 2 of the beam element) at mid-span of the beam

due to a moving sprung load as a function of time

(a)

(b)

(c)


153

4.3.4 Wheel as a moving sprung mass traversing a simply supported beam

In Figure 4.13 a simply supported beam is subjected to a moving sprung mass Mw of

5.75 t traversing a beam at a constant speed from left to right.

Figure 4.13: Simply supported beam subjected to a moving sprung mass

As before, one studies the deflection, vertical acceleration and bending moment at

mid-point of the beam as a function of time in Figures 4.14a, 4.14b and 4.14c,

respectively. The vertical displacement and acceleration of the sprung mass as a

function of time are shown in Figures 4.15a and 4.15b, respectively. In Figure 4.16, a

plot of the contact force between the wheel and the rail is presented. From inspection,

the results from the WRC element are almost identical with the results from ANSYS

CONTAC48 element.

Another interesting feature observed between Figures 4.15b and 4.16 is that vertical

acceleration of the moving sprung mass tends to behave in a somewhat similar

manner to the wheel-rail contact force. As the wheel experiences a positive

acceleration in Figure 4.15b, the wheel rail-contact force undergoes an increase in its

compressive force; whereas, a negative acceleration in Figure 15b tends to decrease

the compressive force.

L


154

-0.0025

-0.0020

-0.0015

-0.0010

-0.0005

0.0000

0.0005

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

WRC Element

ANSYS CONTAC48

-0.4

-0.2

0.0

0.2

0.4

0.6

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al A

cc

ele

rati

on

(m

/s2) WRC Element ANSYS CONTAC48

0

100

200

300

400

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Be

nd

ing

Mo

me

nt

(kN

m)

WRC Element (Element5_LN2)

ANSYS CONTAC48 (Element5_LN2)


moment (local node 2 of the beam element) at mid-span of the beam

due to a moving sprung mass as a function of time

(a)

(b)

(c)


155

-0.0030

-0.0025

-0.0020

-0.0015

-0.0010

-0.0005

0.0000

0.0005

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

WRC Element

ANSYS CONTAC48

-0.20

-0.15

-0.10

-0.05

0.00

0.05

0.10

0.15

0.20

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al A

cc

ele

rati

on

(m

/s2)

WRC Element

ANSYS CONTAC48

Figure 4.15: (a) Vertical displacement; (b) vertical acceleration of the sprung mass

as a function of time

-1.10

-1.05

-1.00

-0.95

-0.90

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Co

nta

ct

forc

e / S

tati

c w

eig

ht

WRC Element

ANSYS CONTAC48

Figure 4.16: Contact force between the sprung mass and beam

(b)

(a)


156

4.3.5 A travelling bouncing wheel traversing a rigid rail and beam

In the following example, one examines the effects of wheel-rail separation, whereby

the wheel is modelled as a travelling sprung mass with an initial positive extension

resulting in a zero contact force. As mentioned earlier, when the extension in the

Hertzian spring becomes positive, no contact exists between the wheel and the rail,

the contact force is zero, and the Hertzian stiffness matrices are put equal to zero.

In this example, rigid rails are located on either side of a simply supported beam as

shown in Figure 4.17. The wheel at time t = 0 sec is at a distance lr of 25m from the

left support of the beam, while the initial vertical extension is 0.01m i.e. separation

between bottom of the wheel and rail is 0.01m. The wheel traverses the simply

supported beam between the times t = 0.9 to 1.8 sec and at all other times the wheel

travels on the rigid rail. The unstrained spring has a length of 0.5m.

Figure 4.17: Rigid rail and simply supported beam subjected to a travelling

bouncing wheel

The vertical displacement of the beam (at mid-point) and the wheel as a function of

time is plotted in Figures 4.18 and 4.19, respectively, while the contact force between

the wheel and rail is shown in Figure 4.20. Again, excellent agreement can be seen

between the solution from using the WRC element and that from using the ANSYS

CONTAC48 element.

L lr


157

-0.006

-0.005

-0.004

-0.003

-0.002

-0.001

0

0.001

0.002

0.003

0.0 0.5 1.0 1.5 2.0 2.5

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

WRC Element

ANSYS CONTAC48

Figure 4.18: Vertical displacement at mid-point of the beam due to the travelling

bouncing wheel

-0.12

-0.10

-0.08

-0.06

-0.04

-0.02

0.00

0.02

0.0 0.5 1.0 1.5 2.0 2.5

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

WRC Element

ANSYS CONTAC48

Figure 4.19: Vertical displacement of the travelling bouncing wheel

-3.0

-2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.0 0.5 1.0 1.5 2.0 2.5

Time (sec)

Co

nta

ct

forc

e / S

tati

c w

eig

ht

WRC Element ANSYS CONTACT48

Figure 4.20: The contact force between the rail and the travelling bouncing wheel

Wheel on

rigid rail

Wheel on

rigid rail


158

One can also see from Figure 4.20 that as the wheel regains contact with the rail it

generates a large impact load, much greater than the weight of the wheel; thus the

maximum deflection of the beam shown in Figure 4.18 is much greater than the

corresponding value in Figure 4.14a.

4.3.6 Sprung & unsprung wheel systems at a wide range of contact stiffness

The WRC element models the wheel of a vehicle as sprung mass involving a Hertzian

spring. In this section, the author compares the results using the WRC element with

the results using the unsprung mass of Olsson (1985); however, the author uses both

hard and soft Hertzian springs for the WRC elements because Hertzian springs are

absent from Olsson’s (1985) model as shown in Figure 4.21.

Figure 4.21: (a) Olsson’s (1985) unsprung wheel; (b) the author’s sprung wheel

The bridge properties used in the following examples are the same as Section 4.3.3,

while dimensionless parameters, similar to those of Olsson (1985), are adopted for the

vehicle, such that the vehicle to bridge mass ratio is 0.5; the unsprung wheel mass to

sprung vehicle mass ratio is 0.25; the bridge to vehicle frequency ratio is 3; the

vehicle damping ratio is 0.125; thus, in this example Mw is 5.75 t, Mv is 23 t, k1 is

(a) Olsson unsprung model (b) Authors’ sprung model


159

2300 kN/m, and c1 is 57.5 kNs/m. Other dimensionless parameters such as speed ratio

α and dynamic amplification factor DAFU have been defined in Section 3.2.3.

In Figure 4.22, the author compares the result using the WRC element with the results

using the unsprung mass of Olsson (1985) under smooth rail conditions for a range of

Hertzian spring stiffness values. The results with the harder Hertzian spring stiffness

have a greater likeness with the solution of Olsson (1985) at all speeds, whereas the

results using softer Hertzian spring stiffness only compare well with Olsson (1985)

results at lower speeds. Hence, the WRC element can also be used for moving

unsprung mass models provided that the Hertzian spring stiffness is given a

reasonably larger value.

1

1.1

1.2

1.3

1.4

1.5

1.6

1.7

0.0 0.2 0.4 0.6 0.8 1.0

Speed ratio α α α α

DA

FU

Hertzian Stiffness = 1.595E3 kN/m



Olsson (1985) Unsprung Mass

Figure 4.22: Simply supported beam subjected to a moving vehicle using different

Hertzian spring stiffness on a smooth rail


160

4.3.7 Sensitivity analysis of the WRC element

It has been shown in Section 3.2.2.3 that the results from the ANSYS contact

elements tend to lose accuracy as the number of beam elements in the model is

reduced; therefore, the author must also examine the WRC element using a coarser

beam mesh, i.e. only 4 beam elements, to see if this is also true for this model. As an

experiment, the author also replaces the shape functions N1 with H1 and N2 with H2, while

G1 and G2 are set to zero, to show that ANSYS contact element is a linear WRC element.

Using the same bridge and vehicle properties as in Section 4.3.4, Figure 4.23 plots the

vertical displacement and bending moment at mid-span of the beam as a function of

time for both the WRC element and ANSYS contact element method. The bridge is

discretized into four beam elements. The vertical displacement of the sprung mass as

a function of time for both systems is presented in Figure 4.24. It can be seen from

each plot that the accuracy of the WRC element is unchanged when the number of

beam elements in the model is reduced from ten to four elements, unlike the ANSYS

contact element system that seems to lose its accuracy. It should be noted that in

Figure 4.23b, the differences in the bending moment for the WRC element at mid-

span relate to the same numerical issue observed in Figure 3.15 and 3.17.

-0.0025

-0.0020

-0.0015

-0.0010

-0.0005

0.0000

0.0005

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

WRC Element - 4 elements

ANSYS CONTAC48 - 4 elements

Linear WRC Element - 4 element

Figure 4.14a - 10 elements

(a)


161

0

100

200

300

400

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Be

nd

ing

Mo

me

nt

(kN

m)



Linear WRC Element - 4 elements

Figure 4.14c - 10 elements

Figure 4.23: (a) Vertical displacement; (b) bending moment (local node 2 of the

beam element) at mid-span of the beam due to a moving sprung mass

as a function of time using only 4 beam elements

-0.0030

-0.0025

-0.0020

-0.0015

-0.0010

-0.0005

0.0000

0.0005

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)



Linear WRC Element - 4 elements

Figure 4.15a - 10 elements

Figure 4.24: Vertical displacement of the sprung mass as a function of time using

only 4 beam elements

Another issue that the author wishes to examine during this sensitivity analysis of the

WRC element relates to a moving sprung mass traversing a cantilever beam. In

Section 4.3.2, it is shown that the moving sprung mass loses contact with a cantilever

beam, whichever direction it is travelling in; therefore, the author considers modelling

both cantilever beams again using 2000 equal time-step to ensure that the solution is

(b)


162

unchanged. The reader can see the contact force that exists between the mass and

beam for both the fixed-free and free-fixed cantilever beam in shown Figures 4.25a

and 4.25b, respectively. It can be observed from the results that there are minimal

differences between the graphs of Figure 4.25 and Figure 4.9. The latter only uses 200

equal time-steps. Wheel-rail separation still occurs in both solutions.

-2.0

-1.5

-1.0

-0.5

0.0

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Co

nta

ct

forc

e / S

tati

c w

eig

ht

WRC Element (200 ts)


-16

-14

-12

-10

-8

-6

-4

-2

0

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Co

nta

ct

forc

e / S

tati

c w

eig

ht



Figure 4.25: Contact force between the wheel and rail using 2000 equal time-steps:

(a) fixed-free beam; (b) free-fixed beam

4.3.8 Simply supported beam subjected to a two-wheeled sprung system

In this section, the author examines the effects of a two-wheeled sprung system

traversing a simply supported beam as shown in Figure 4.26. The vehicle comprises a

rigid vehicle body supported by a pair of axles by means of primary suspensions,

(a)

(b)


163

consisting of a spring and dashpot. Like the bridge and vehicle of Yang and Wu

(2001), the bridge has a length L of 30m, Young’s modulus of elasticity E of

2.943x107 kN/m


4, mass per unit length m of 36 t/m

and a Poisson’s ratio ν of 0.2, while the vehicle model has two sprung wheel masses

Mw of 0 t, a vehicle mass Mv of 540 t, a rotational moment of inertia Iv of 13800 t m2,

suspension stiffness k1y of 41350 kN/m, suspension damping c1y of 0 kN s/m, distance

between wheel axles Aw of 17.5 m and a vehicle speed c of 27.778 m/s (100 km/hr).

The Hertzian spring stiffness adopted is kH = 1.595 x107 kN/m. In this model, the

mass and moment of inertia of the vehicle is applied at the centre node of the vehicle

body (Vc) by means of two massless beam elements supported by the suspension springs.

Longitudinal springs with a stiffness k1x of 1000 kN/m are also included in this model.

Figure 4.26: Two-wheeled sprung system traversing a simply supported beam

A plot of the vertical displacement and acceleration at mid-span of the beam as a

function of time can be seen in Figures 4.27a and 4.27b, respectively, while the

vertical acceleration of the vehicle body (Vc) is presented in Figure 4.28. The wheel-

rail contact force for the front wheel of the vehicle is shown in Figure 4.29. As before,

the results from using the WRC element are similar to those from using the ANSYS

contact element, provided that a sufficient number of beam elements are used in the

model.

wM

wM

1yc 1y

c1y

k 1yk

Hk

, v v

M I

cc

V

L

, , E I mwA

Hk

1xk1x

k


164

-0.010

-0.008

-0.006

-0.004

-0.002

0.000

0.002

0.004

0.0 0.5 1.0 1.5 2.0


Ve

rtic

al D

isp

lac

em

en

t (m

)

WRC Element

ANSYS CONTAC48

-3.0

-2.0

-1.0

0.0

1.0

2.0

3.0

0.0 0.5 1.0 1.5 2.0


Ve

rtic

al A

cc

ele

rati

on

(m

/s2)

WRC Element

ANSYS CONTAC48

Figure 4.27: (a) Vertical displacement; (b) vertical acceleration at mid-span of the

beam as a function of time due to a two-wheeled vehicle

-0.50

-0.25

0.00

0.25

0.50

0.0 0.5 1.0 1.5 2.0


Ve

rtic

al A

cc

ele

rati

on

(m

/s2)

WRC Element

ANSYS CONTAC48

Figure 4.28: Vertical acceleration of the vehicle body (Vc) of a two-wheeled vehicle

as a function of time

(a)

(b)


165

-1.10

-1.05

-1.00

-0.95

-0.90

0.0 0.5 1.0 1.5 2.0


Co

nta

ct

forc

e / S

tati

c w

eig

ht

WRC Element

ANSYS CONTAC48

Figure 4.29: The contact force between the rail and front wheel of the vehicle as a

function of time

4.3.9 Two-wheeled vehicle subjected to braking effects

Until now, one has only focused on the vertical response of the bridge and vehicle;

however, in this section one examines both the vertical and horizontal responses of a

two-wheeled vehicle as it is decelerating along the bridge. The bridge and vehicle

properties are the same as those of Section 4.3.8. In all examples, as the front wheel of

the vehicle enters the bridge at time t = 0 sec, it decelerates at a constant value a

(positive, if accelerating). The distance the vehicle travels as it decelerates is 60 m.

Like Yang and Wu (2001), one considers the following three cases: (1) 50c = m/s,

10a = − m/s2; (2) c = 100 m/s, 10a = − m/s

2; (3) 100c = m/s, 20a = − m/s

2. On this

occasion, results from the developed models are compared with the results from the

literature, as the ANSYS contact elements are unable to simulate the braking effects

of the vehicle. With the aid of Equation (G.10), the author specifically computes the

times when the front wheel of the vehicle arrives on the left-hand support, mid-span

and right-hand support of the beam. In addition, when the front wheel has travelled

47.5 m (Aw + L), the rear wheel of the vehicle arrives on the right hand support. These

computed times are presented in Tables 4.1 to 4.3 for the three models examined.


166

Table 4.1: Model 1 - Vehicle decelerating at -10 m/s2


Train decelerates from 100.000 m/s to 93.808 m/s over 60 m

Time X(t) Vel(t) Comments

(s) (m) (m/s)

0.000 0.000 100.000 Front wheel of vehicle at LHS

0.151 15.000 98.488 Front wheel of vehicle at mid-span

0.305 30.000 96.954 Front wheel of vehicle at RHS

0.487 47.500 95.132 Rear wheel of vehicle at RHS




(s) (m) (m/s)





It should it noted that the speed c used in the dimensionless parameter ct/L in the

following results is computed by averaging the initial and final speed of the vehicle in

each model. For comparison purposes, the vertical displacement at mid-span of the

beam and vertical acceleration of the vehicle body as a function of time are compared

with the results from the literature. The author’s results are shown in Figures 4.30a

and 4.31a, while Yang and Wu (2001) results are shown in Figures 4.30b and 4.31b.



(s) (m) (m/s)






167

From inspection of these results, it can be seen that there are good similarities

between the author’s solutions and Yang & Wu’s (2001) solutions. Both sets of

results display similar shapes and magnitudes, especially the 2nd

and 3rd

model. In

addition, the results show that the 2nd

and 3rd

model tend to give a larger vertical

response than the 1st model, which indicates that the speed of the vehicle entering the

bridge, at time t = 0 sec, has a greater influence on the vertical response than the

braking effect of the vehicle, even though the deceleration of the vehicle occurs over a

quite limited time frame according to Yang & Wu (2001).

-0.010

-0.008

-0.006

-0.004

-0.002

0.000

0.002

0.004

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0


Ve

rtic

al D

isp

lac

em

en

t (m

)

WRC (Model1)

WRC (Model2)

WRC (Model3)

Figure 4.30: Vertical displacement at mid-span of the beam as a function of time

due to vehicles decelerating: (a) numerical solution; (b) Yang and Wu

(2001) solution

(a)


(b) Yang & Wu (2001) Model 1

Yang & Wu (2001) Model 2

Yang & Wu (2001) Model 3


168

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0


Ve

rtic

al A

cc

ele

rati

on

(m

/s2)

WRC (Model1)

WRC (Model2)

WRC (Model3)

Figure 4.31: Vertical acceleration of the vehicle body as a function of time due to

vehicles decelerating: (a) numerical solution; (b) Yang and Wu (2001)

solution

A plot of the horizontal reaction force experienced by the pinned support as the

vehicle decelerates at the different speeds is shown in Figure 4.32. It can be seen from

Figure 4.32 that the horizontal reaction force tends to increase significantly as each

wheel enters the bridge and then as each wheel leaves the bridge there is a sudden

drop in the force followed by a fluctuation of the force, which is a similar observation

made in the Yang & Wu (2001) study. It is unusual from Yang & Wu (2001) results

that all three models tend to give the same shape and values especially when

Newton’s Second Law of motion states that F = Ma and secondly their magnitude

differs by a factor of 5 (Model 1 and 2) and 10 (Model 3) with the author’s solution.

(b)

(a)


Yang & Wu (2001) Model 1 (c = 50 m/s; a = -10 m/s^2)




169

-15000

-10000

-5000

0

5000

10000

15000

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0


Re

ac

tio

n F

orc

e (

kN

)

Model 1 (WRC)

Model 2 (WRC)

Model 3 (WRC)

Figure 4.32: Horizontal reaction force due to the vehicles decelerating: (a) WRC

numerical solution; (b) Yang and Wu (2001) solution

Furthermore, Yang & Wu (2001) relates the ith wheel horizontal contact force i

H to

its vertical weight i

V by i i i

H Vµ= , where i

µ is the friction coefficient for the ith

wheel. This implies that their horizontal reaction force is not dependent on the

acceleration of the vehicle unlike in the author’s system; hence, all three of their

results have similar magnitude and shape. Inspecting Figure 4.32a, the reader can see

that the author’s results are dependant on the acceleration of the vehicle. The bridge

reaction force is then minus one times the applied force.




(a)

(b)


170

4.4 Application of WRC element to Boyne Viaduct

Based on the results of Section 4.3, the author can claim that the wheel-rail contact

element is exceedingly robust for smooth rail conditions; hence, one is in a position to

apply this technique to the Boyne Viaduct railway-bridge subjected to a moving

vehicle. In this section, one begins by representing the railway vehicle as a single

moving sprung mass, while later on in the section; a realistic three-dimensional

vehicle comprising wheels, suspensions and vehicle bodies is used to represent the

train. Depending on the results sought, one can adopt a simple or complex vehicle

model. Should one be primarily interested in the dynamic response of the bridge, then

the single moving sprung mass is the better option. However, if one is also interested

in the vehicle response then a two-dimensional or three-dimensional model of the

vehicle is needed. The two-dimensional bridge and vehicle model has the advantage

of being less complex than the three-dimensional bridge and train model; however,

one is only able to obtain displacements in the vertical and horizontal directions as

well as the rotation about the z-axis (equivalent to the pitching motion for the vehicle)

for the former model. In the three-dimensional bridge and train model, one can obtain

displacements in all three directions (vertical, horizontal and lateral) as well as

obtaining the three rotations about the x, y and z-axes (equivalent to the rolling,

yawing and pitching motion of the vehicle) as external loads are applied to the train.

4.4.1 Single sprung wheel traversing two-dimensional Boyne Viaduct

One starts by examining a two-dimensional Boyne Viaduct railway bridge subject to a

single moving sprung wheel travelling at slow (10 km/hr) and fast speed (164 km/hr).

Since each wheel from the 201 Class Irish-Rail locomotive has a weight of 91.25 kN

as shown in Figure 3.31, its mass Mw is equal to 9.3 t. In this example, 0.1P G ≈ and

the Hertzian contact stiffness used throughout is kH = 1.4 x 106 kN/m, while time t is


171

arranged in such a manner that the wheel arrives on the bridge at t = 0 sec; thus,

exiting the bridge at 1ct L = . It is unrealistic to model the whole train as a single

moving sprung mass; instead, only the front wheel of the train is chosen in this model.

In Figure 4.33, the author plots the vertical displacement and internal axial forces at

mid-span of the bridge as a function of time. As before, the static effects of the bridge

are omitted; hence, only the dynamic effects are taken into account. From inspection

of Figure 4.33, the reader can see that the dynamic response of the bridge is minimal

at the slow speed (10km/hr), while at the higher speed (164 km/hr) the bridge tends to

experience a noticeable increase in deflection and axial force.

In addition, one also presents the vertical acceleration of the vehicle i.e. the sprung

wheel, which is often used as a measurement of the riding comfort experienced by the

passengers. Many studies such as those of Yau et al. (1999) and Wu and Yang (2003)

recommend that the vertical acceleration not exceed 0.5 m/s2 (as adopted by France-

SNCF and Taiwan-HSR) while the Eurocodes (1990) use a less severe value of 1.0

m/s2. In any case, prolonged exposure to accelerations exceeding these recommended

values can lead to a reduction in the passenger riding comfort. It can be seen from

Figure 4.34 that at the slow speed (10 km/hr) the riding comfort value for sprung mass

wheel lies well within the recommended values, while at the faster speed of 164 km/hr

the sprung mass wheel tends to exceed the value of 0.5 m/s2 on numerous occasions,

whilst remaining within the Eurocodes (1990) value of 1.0 m/s2.

In Figure 4.35 the contact force that exists between the moving sprung wheel and

bridge is plotted. As observed in previous results, the contact force effects between

the vehicle and bridge are minimal at the slow speed and quite substantial at the faster

speed. Nevertheless, wheel-rail separation is not present for this particular problem.


172

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.2 0.4 0.6 0.8 1.0


Dy

na

mic

co

eff

icie

nt

Moving Sprung Wheel - 10 km/hr


-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

0.0 0.2 0.4 0.6 0.8 1.0


Ax

ial fo

rce

co

eff

icie

nt



Figure 4.33: (a) Vertical displacement; (b) axial force at mid-span of the Boyne

Viaduct due to a sprung wheel travelling at 10 km/hr and 164 km/hr

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

0.0 0.2 0.4 0.6 0.8 1.0


Ve

rtic

al a

cc

ele

rati

on

(m

/s2)

Moving Sprung Wheel - 10 km/hr (multiplied by a factor of 50)


Figure 4.34: Vertical acceleration of the wheel i.e. riding comfort

(a)

Bottom Chord

Top Chord

(b)


173

-1.15

-1.10

-1.05

-1.00

-0.95

-0.90

0.0 0.2 0.4 0.6 0.8 1.0


Co

nta

ct

forc

e / S

tati

c w

eig

ht



Figure 4.35: Contact force between the wheel and bridge as a function of time

For a more in-depth analysis of the Boyne Viaduct subjected to a moving sprung

wheel, the author now considers the dynamic response of the bridge at a range of

wheel speeds. From Section 3.2.3, the critical speed of the vehicle has already been

computed as 550 m/s,cr

c = or 1980 km/hr. However, it should be noted that a more

realistic range of speeds between 10 km/hr to 300 km/hr i.e. 0 0.15α< < are of

greater interest to the author. Furthermore, it is observed from Figure 4.36 that at

extremely high speeds >300 km/hr, the sprung wheel tends to lose contact with the

bridge and the impact loads experienced by the bridge as the wheel regained contact,

greatly increases the dynamic amplification factor as shown in Figure 4.37. The

contact force between the sprung wheel and bridge at speed ratios

0.125 and 0.25,α = which is equivalent to 247 km/hr and 495 km/hr is shown in

Figure 4.36. This plot shows that wheel-rail separation occurs frequently at the

excessively larger speed. The dynamic amplification factors DAFU and DAFA for the

sprung wheel at a range of speeds can be seen in Figure 4.37. As the speed of the

sprung wheel increases beyond 0.25,α < the impact loading due to wheel-rail

separation gives very large DAFU and DAFA values.


174

-5.0

-4.0

-3.0

-2.0

-1.0

0.0

0.0 0.2 0.4 0.6 0.8 1.0


Co

nta

ct

forc

e / S

tati

c w

eig

ht



Figure 4.36: Contact force between the wheel and bridge at extremely high speeds

1.0

1.2

1.4

1.6

1.8

2.0

0.0 0.2 0.4 0.6 0.8 1.0


DA

FU

Single Sprung Wheel (P/G = 0.1)

Moving Load - Figure 3.23a

1.0

1.5

2.0

2.5

3.0

3.5

4.0

0.0 0.2 0.4 0.6 0.8 1.0


DA

FA

Single Sprung Wheel (P/G = 0.1) - Bot Chord

Single Sprung Wheel (P/G = 0.1) - Top Chord

Moving Load - Bot Chord - Figure 3.23b

Moving Load - Top Chord - Figure 3.23b

Figure 4.37: Dynamic amplification factor at mid-span of the Boyne Viaduct versus

speed ratio: (a) deflection; (b) axial internal forces

(a)

(b)


175

A closer examination of the DAFU at a range of realistic speed values, as represented

by the vertical dashed line in Figure 4.37a, can be seen in Figure 4.38. It can be

observed from Figure 4.38 that the additional wheel inertia tends to slightly increase

the dynamic response of the bridge. Furthermore, one can see that as the speed of the

sprung wheel passes between 80 km/hr to 100 km/hr, the DAFU is significantly

greater than the moving load, while at most other speeds the difference in the DAFU

between the moving load and sprung wheel is somewhat less.

1.00

1.05

1.10

1.15

1.20

0 50 100 150 200 250 300

Speed (km/hr)

DA

FU

Single Sprung Wheel (P/G = 0.1)

Moving Load - Figure 3.24

Figure 4.38: A close-up view of Figure 4.37a using a realistic range of speeds

4.4.2 Railway Vehicle

As shown in Figure 4.39, each vehicle consists of a vehicle body supported by a pair

of bogies, with each bogie supported by axles. Finally a pair of wheels supports each

axle. The bogies are connected to the axles through primary suspensions and to the

vehicle body through secondary suspensions, with each suspension consisting of a

spring and dashpot. In the two-dimensional models, one assumes that the weight of all

components is halved.


176

secondary suspension

primary suspension

Hertian spring

vehicle body

bogie

wheelset

vehicle body

bogie

Side View

bogie

Front View

wheels wheels


primary suspension

Hertian spring

vehicle body

bogie

wheelset

vehicle body

bogie

Side View

bogie

Front View

wheels wheels

Figure 4.39: Three-dimensional railway carriage model (a) descriptive; (b) using

symbolization

The total mass of each wheel, bogie frame and vehicle body is denoted by the

symbols Mw, Mb and Mv. In Appendix G, the author computes the equivalent mass

moment of inertia for the bogie frame and vehicle body and these values can be found

in Table 4.4. The primary spring stiffness and damping is given by k1 and c1, while k2

and c2 denote the secondary spring stiffness and damping. The Hertzian spring

stiffness, also known as the wheel-rail contact stiffness, is given by kH. Longitudinal

and lateral springs are required between the vehicle body and the bogie and also

between the bogie and axles to prevent the structure from becoming a mechanism.

These horizontal springs are independent from the main suspension springs.

(a)

2 yk 2 y

c

1yk 1y

k

2 yk

1yc

1yc

2 yc

HkH

kw

Mw

Mw

Mw

M

,v zzv

M I

,b zzb

M I

(b)

,b zzb

M I

,v xxv

M I

bM

1 1,z z

k c

1 1,y y

k c

1 1,z z

k c

1 1,y y

k c

Front View

Side View

2wA

1wA

wb

vb

vl

0h

1h

3h

y

x

y

z

4wA

3wA

wB

2 2,x x

k c

vh


177

Each wheel of the train is modelled using lumped masses, while elastic beam

elements are used to model the bogie frame and vehicle body. The primary and

secondary suspensions are then modelled using spring elements. Figure 4.40 shows

the author’s six-axle 201 Class locomotive and four-axle Mark3 railway coach

modelled in ANSYS. In addition, the author indicates, in particular, the 1st (W1 is the

front wheel of the locomotive) and 7th

(W7 is the front wheel of the first railway

coach) wheel of the train. The nodes V1 and V2 on the vehicle bodies of the

locomotive and railway coach, respectively, are also shown. These nodes are

examined in the results.

Figure 4.40: Six-axle 201 Class locomotive and four-axle Mark3 railway coach

modelled using beam, spring and lumped mass elements in ANSYS

(Bowe and Mullarkey, 2005)

A list of vehicle properties and dimensions for both the Class 201 locomotive and

Mark3 railway coach adopted in this study are presented in Table 4.4. It should be

noted that the symbol X marked in the left hand column, indicates that this particular

value is not used or required in the two-dimensional vehicle model.

V1

W1 – First wheel of the train (node)

W7 – Seventh wheel of the train (node)

V1 – Vehicle body of the locomotive (node)

V2 – Vehicle body of the railway coach (node) V2

W1

W7


178

Table 4.4: Vehicle dimensions and parameters (Irish Rail vehicle data sheet)

Data Symbols Unit Loco 201 MK3

coach

Mass properties

Mass of vehicle body Mv t 64.48 35.74

x Roll inertia of vehicle body Ixxv tm2 123.0 68.2

x Yaw inertia of vehicle body Iyyv tm2 2008 1287

Pitch inertia of vehicle body Izzv tm2 2002 1284

Mass of bogie frame Mb t 10.18 3.15

x Roll inertia of bogie frame Ixxb tm2 10.18 3.15

x Yaw inertia of bogie frame Iyyb tm2 21.73 4.91

Pitch inertia of bogie frame Izzb tm2 11.55 1.76

Mass of wheel Mw t 4.52 1.50

Roll inertia of wheel Ixxw tm2 4.52 1.50

Suspension stiffness

Primary suspension in the longitudinal direction k1x kN/m 4240 20260

Secondary suspension in the longitudinal direction k2x kN/m 320 422

Primary suspension in the vertical direction k1y kN/m 1470 3185

Secondary suspension in the vertical direction k2y kN/m 630 566

x Primary suspension in the lateral direction k1z kN/m 2120 10130

x Secondary suspension in the lateral direction k2z kN/m 160 211

Suspension damping

Primary suspension in the longitudinal direction c1x kNs/m 1.00 0.000

Secondary suspension in the longitudinal direction c2x kNs/m 32.00 41.44

Primary suspension in the vertical direction c1y kNs/m 4.00 32.41

Secondary suspension in the vertical direction c2y kNs/m 20.00 26.24

x Primary suspension in the lateral direction c1z kNs/m 1.00 0.00

x Secondary suspension in the lateral direction c2z kNs/m 32.00 41.44

Dimension - longitudinal direction - x

1st axle to c.o.g. of bogie frame Aw1 m 1.689 1.300

2nd axle to c.o.g. of bogie frame Aw2 m 0.000 1.300

3rd axle to c.o.g. of bogie frame Aw3 m 2.019 1.300

4th axle to c.o.g. of bogie frame Aw4 m 1.689 1.300

5th axle to c.o.g. of bogie frame Aw5 m 0.000 -


Rear axle of a vehicle to front axle of next vehicle Cw m 4.169 4.400

Front bogie frame c.o.g. to c.o.g. of rear bogie frame Bw m 13.405 16.000

Half of bogie frame to c.o.g. of vehicle body lbg m 6.703 8.000

Overall length of vehicle body lv m 19.113 20.600

Dimension - vertical direction - y

Height from rail to wheel centre h0 m 0.508 0.460

Height from rail to c.o.g of bogie h1 m 0.843 0.600

Height from rail to secondary suspension h2 m 0.843 0.600

Height from rail to c.o.g. of vehicle body h3 m 1.393 0.220

Overall height of vehicle body hv m 2.200 2.200

Dimension - lateral direction - z

x Gauge width of wheelset bw m 1.500 1.500

x Overall width of vehicle body bv m 2.000 2.000


179

4.4.3 Two-dimensional bridge-train model

In this section, the author analyses the vertical response of the two-dimensional

Boyne Viaduct subjected to the moving train that is described in Section 4.4.2. As

with previous models, one examines the dynamic response of the bridge subjected to a

vehicle traversing the bridge at a slow speed (10 km/hr) as well as at a high speed

(164 km/hr). Time t is arranged in such a manner that the front wheel enters the bridge

at time t = 0 sec and exits the bridge at 1ct L = . It should also be worth noting that

for the following example the vehicle-to-bridge ratio is computed as P/G = 1.34;

however, this value doesn’t take into account that the train, which measures a length

of 85.88 m from the front wheel to the rear wheel of the train, is slightly longer than

the bridge,. Therefore, the P/G ratio experienced by the bridge varies from 1.20 to

1.05, which relates to the front wheel of the train arriving on the right-hand support

and the rear wheel of the train arriving on the left-hand support, respectively.

A plot of the vertical displacement and axial force in the top and bottom chord at mid-

span of the two-dimensional Boyne Viaduct subjected to a moving train at the two

different speeds is shown in Figure 4.41. It can be seen that the results for faster speed

(164 km/hr) have a greater oscillatory motion than the results for slower speed (10

km/hr); however, the oscillations of the response are smaller than the oscillations of

the moving load solution in Figure 3.33. In addition to the dynamic response of the

bridge, one also examines the vertical acceleration or riding comfort value of the train

travelling at the faster speed (164 km/hr) as shown in Figure 4.42. This plot shows

that the vertical accelerations of both the locomotive (V1 of Figure 4.40) and 1st

railway coach (V2 of Figure 4.40) lie well within the recommended value of 1.0 m/s2

specified in Eurocodes (1990), for the higher vehicle speed.


180

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0


De

fle

cti

on

co

eff

icie

nt

2D Boyne - 2D Train at 10 km/hr


-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

0.0 0.5 1.0 1.5 2.0


Ax

ial fo

rce

co

eff

icie

nt



Figure 4.41: (a) Vertical displacement; (a) axial force at mid-span of the 2D Boyne


-1.50

-1.00

-0.50

0.00

0.50

1.00

1.50

0.0 0.5 1.0 1.5 2.0


Ve

rtic

al A

cc

ele

rati

on

(m

/s2)

2D Train - 164 km/hr (V1) 2D Train - 164 km/hr (V2)

Figure 4.42: Vertical acceleration (riding comfort value) of the vehicle bodies

(a)

(b)

Bottom chord

Top chord


181

In Figure 4.43, the contact force that exists between the 1st and 7

th wheel of the train

and the rail are examined at the faster speed (164 km/hr). It can be seen from the

results that, as wheels enter the bridge or immediately after they leave the bridge

(arrives on the rigid rail); the compressive force between the wheel and rail tends to

oscillate abruptly, gradually reducing in amplitude over time.

-1.4

-1.2

-1.0

-0.8

-0.6

0.0 0.5 1.0 1.5 2.0


Co

nta

ct

forc

e / S

tati

c w

eig

ht

2D Train - 164 km/hr (W1)


Figure 4.43: Contact force between the 1st and 7

th wheel and the rail

Next, the author examines the dynamic response of the two-dimensional Boyne

Viaduct subjected to the moving train at a range of realistic speeds between

0 0.15α< < i.e. 10 km/hr to 300 km/hr. Exceeding this particular range of speed is

likely to increase the chances of wheel-rail separation, which can greatly change the

dynamic behaviour of the bridge, as shown in Section 4.3.5. Figure 4.44 presents the

dynamic amplification factors DAFU and DAFA for a range of speeds between 10

km/hr to 300 km/hr. First, it can be seen from Figure 4.44 that results from the

author’s wheel-rail contact (WRC) element compare well with the results from

ANSYS contact elements, while second, it is observed that the results for moving

two-dimensional train model are much less oscillatory than results for the bridge

subjected to a moving load.


182

1.00

1.02

1.04

1.06

1.08

1.10

0 50 100 150 200 250 300

Speed (km/hr)

DA

FU

2D Boyne (2D Train using WRC)

2D Boyne (2D Train using ANSYSCE)

2D Boyne (Moving Load)

1.00

1.02

1.04

1.06

1.08

1.10

0 50 100 150 200 250 300

Speed (km/hr)

DA

FA

Bot Chord (2D Train using WRC)

Bot Chord (2D Train using ANSYSCE)

Bot Chord (Moving Load)

1.00

1.02

1.04

1.06

1.08

1.10

0 50 100 150 200 250 300

Speed (km/hr)

DA

FA

Top Chord (2D Train using WRC)

Top Chord (2D Train using ANSYSCE)

Top Chord (Moving Load)


versus Speed (a) vertical displacement; (b) axial force in the bottom

chord; (c) axial force in the top chord.

(a)

(b)

(c)


183

4.4.4 Braking and accelerating effects of train on the bridge

Until now, it has been assumed the train has a constant speed as it travels across the

bridge; however, utilizing the longitudinal spring elements described in Section 4.2.2,

the author is also able to simulate the braking and accelerating effects of a vehicle as

it travels along the bridge. This simulation is not possible with the ANSYS contact

element. In the following two examples, the author simulates the braking and then the

accelerating effects of a train as it crosses the Boyne Bridge. It should be noted that as

the train brakes, all the wheels of the train experience a horizontal longitudinal force

from the rail; whereas, when the train accelerates, only the wheels of the locomotive

experience the horizontal longitudinal force from the rail.

In the first two examples, it is assumed that the train is given an initial speed c of

45.55 m/s (164 km/hr) at time t = 0 sec and the moment the front wheel enters the

bridge it begins to decelerate at: (1) a = -2.075 m/s2 so that the train comes to a

complete stop after 500 m; (2) a = -1.037 m/s2 so that the train comes to a complete

stop after 1000 m (the velocity and wheel position at any time are computed using

Equation G.10). These decelerations are likely to occur when the emergency brakes are

applied to the train (Wikipedia, 2010). In the next two examples, it is then assumed

that the train begins from a stationary position, such that the front wheel is on the left

hand support at time t = 0 sec and begins to accelerate at: (3) a = 2.075 m/s2 so that

the train reaches a speed of 45.55 m/s after 500 m; (4) a = 1.037 m/s2 so that the train

reaches a speed of 45.55 m/s after 1000 m. A plot of the position of the front wheel of

the train versus the train speed is shown in Figure 4.45. The two vertical dashed lines,

shown on each plot, indicate the front wheel and rear wheel of the train arriving on the

right hand support of the bridge i.e. front wheel travels 80.77 m and 166.65 m (see

Section 3.3.3.1 for the distance between the front and rear wheel of the train used).


184

Figure 4.45: Position of the front wheel of the train versus the train speed (a) train

decelerating; (b) train accelerating

With the aid of Equation (G.10), the author has specifically computed the times when

the front wheel of the train arrives on the left-hand support, mid-span and right-hand

support of the bridge. In addition, when the front wheel has travelled 166.65 m, the

rear wheel of the train (single 201 Class locomotive and three Mark3 railway coaches)

arrives on the right hand support. These computed times are presented in Tables 4.5 to

4.8 for the four models examined.

(a)

(b)


185

Table 4.5: Model 1 - Train decelerating at -2.075 m/s2

Table 4.6: Model 2 - Train decelerating at -1.037 m/s2

Train decelerates from 45.55 m/s to 0 m/s over 1000 m


(s) (m) (m/s)

0.000 0.000 45.550 Front wheel of train at LHS

0.896 40.385 44.621 Front wheel of train at mid-span

1.811 80.770 43.672 Front wheel of train at RHS

3.825 166.650 41.582 Rear wheel of train at RHS

Table 4.7: Model 3 - Train accelerating at -2.075 m/s2

Train accelerates from 0 m/s to 45.55 m/s over 500 m


(s) (m) (m/s)





Table 4.8: Model 4 - Train accelerating at -1.037 m/s2



(s) (m) (m/s)







(s) (m) (m/s)






186

The vertical displacement at mid-span and horizontal reaction force in the left hand

support of the bridge as a function of time for a train traversing the two-dimensional

Boyne Bridge are shown in Figure 4.46 and 4.47, respectively. For comparison

purposes, the results from the braking train (models 1 and 2) are plotted on the same

graph, while the results for the train accelerating (models 3 and 4) are also plotted on

the same graph. It should it noted that the speed c used in the dimensionless parameter

ct/L is computed by averaging the speeds given in Tables 4.5 to 4.8 for the front wheel.

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4


Dy

na

mic

co

eff

icie

nt

Model 1 - Deceleration = -2.075 m/s^2


-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0


Dy

na

mic

co

eff

icie

nt

Model 3 - Acceleration = 2.075 m/s^2 Model 4 - Acceleration = 1.037 m/s^2

Figure 4.46: Vertical displacement at mid-span of the 2D Boyne Viaduct due to

moving train (a) decelerating; (b) accelerating

(a)

(b)


187

-100

0

100

200

300

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4


Re

ac

tio

n F

orc

e (

kN

)



-150

-100

-50

0

50

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0


Re

ac

tio

n F

orc

e (

kN

)

Model 3 - Acceleration = 2.075 m/s^2

Model 4 - Acceleration = 1.037 m/s^2

Figure 4.47: Horizontal reaction force in the left hand support of the 2D Boyne

Viaduct due to moving train (a) decelerating; (b) accelerating

From inspection of the results, it can be seen that the vertical response of the bridge as

the train accelerates tends to be slightly greater than, the braking train, at these

particular accelerations. The horizontal response of the bridge tends to be in good

agreement with Section 4.3.9, that is, the reaction force increases significantly as each


drop in the force followed by a fluctuation of the results, until all wheels of the train

have exited the bridge and the reaction force oscillates about zero.

(a)

(b)


188

4.4.5 Three-dimensional bridge-train model

The author now examines the dynamic response of the Boyne Viaduct subjected to a

three-dimensional vehicle initially at operational speeds of 10 km/hr and 164 km/hr,

and later for a wide range of speeds. Again for comparison purposes, one compares

these results with the results for the moving loads of Section 3.3.5.3.

In Figure 4.48, a plot of the vertical displacement, axial force in the upper and lower

chord and bending moment in the mid-span of the cross-beam located at mid-span of

the bridge can be seen for the two different train speeds. Comparing the results in

Figure 4.48 with results in Figure 4.41, the reader can see that the two-dimensional

bridge response is remarkably similar to the three-dimensional bridge response.

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0


De

fle

cti

on

co

eff

icie

nt



Figure 4.41a - 2D Train at 164 km/hr

-1.2

-0.9

-0.6

-0.3

0.0

0.3

0.6

0.9

1.2

1.5

0.0 0.5 1.0 1.5 2.0


Ax

ial fo

rce

co

eff

icie

nt



Figure 4.41b - 2D Train at 164 km/hr

(a)

(b)

Bottom chord

Top chord


189

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0


Be

nd

ing

mo

me

nt

co

eff

icie

nt 3D Boyne - 3D Train at 10 km/hr


Figure 4.48: (a) Vertical displacement; (b) axial force; (c) bending moment of the

cross-beam located at mid-span of the 3D Boyne Viaduct due to a

three-dimensional vehicle traversing at 10 km/hr and 164 km/hr

Next, the author examines the vertical acceleration of the vehicle as it travels as

shown in Figure 4.49. From inspection of Figure 4.49, it can be seen that the vertical

acceleration or riding comfort of the locomotive (V1) and first railway coach (V2) lie

within the recommended value of 1.0 m/s2, on most occasions, as required by

Eurocodes (1990). Figure 4.49 shows that the three-dimensional locomotive (V1)

tends to be slightly more responsive than its two-dimensional counterpart, while the

responses of both types of railway coaches models (V2) behave quite similarly.

-1.50

-1.00

-0.50

0.00

0.50

1.00

0.0 0.5 1.0 1.5 2.0


Ve

rtic

al A

cc

ele

rati

on

(m

/s2)



Figure 4.49: Vertical acceleration (riding comfort value) of the vehicle bodies

(c)


190

The contact forces between the front wheel of the locomotive and the rail and the

front wheel of the first railway coach i.e. the 1st and 7

th wheel of the train and the rail

are shown in Figure 4.50. Like the contact force in the two-dimensional counterpart,

as each wheel enter the bridge or immediately after they leave the bridge (arrives on

the rigid rail), the contact force between the wheel and rail tends to oscillate abruptly,

gradually reducing in time. In addition, it can be seen that the front wheel of the first

railway coach i.e. the seventh wheel of the train, loses contact with the rail as it enters

the bridge. Nonetheless, wheel-rail separation for the 7th

wheel is brief and the contact

force grows as this wheel travels along the bridge. Figure 4.50 shows that wheel-rail

separation of the railway carriage wheels can be more susceptible at higher speed

when entering a bridge structure because the bridge can be experiencing free vibration

or periodic oscillations generated by other wheels already on the bridge.

-2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.0 0.5 1.0 1.5 2.0


Co

nta

ct

forc

e / S

tati

c w

eig

tht





To finish, the author investigates the dynamic response of the three-dimensional

Boyne Bridge subjected to a locomotive and several railway carriages travelling at a

realistic range of speeds, 0 0.165,α< < which is equivalent to 0 km/hr to 300 km/hr.


191

The dynamic amplification factors are presented Figure 4.51. It can be seen from

Figure 4.51a that the dynamic amplification factors DAFU and DAFA of the three-

dimensional structure are quite similar to that of its two-dimensional counter-part at

lower speeds, varying somewhat at higher speeds. In addition, it can be seen that the

dynamic behaviour of the bridge tends to be significantly larger for the moving forces

when compared to the bridge response for two-dimensional and three-dimensional

vehicle models.

1.00

1.02

1.04

1.06

1.08

1.10

1.12

0 50 100 150 200 250 300

Speed (km/hr)

DA

FU



3D Boyne (Moving Load)

1.00

1.02

1.04

1.06

1.08

1.10

0 50 100 150 200 250 300

Speed (km/hr)

DA

FU






versus speed ratio: (a) vertical displacement; (b) axial force

(a)

(b)


192

4.4.6 Effects of lateral cross winds on the train as it travels

As a final section, the author utilizes the lateral spring element described in Section

4.2.3 by briefly examining the effects of a lateral cross wind on a moving train. This

section demonstrates the effectiveness of the lateral spring elements. However, the

aerodynamic action of the wind as a train travels goes beyond the scope of this thesis.

Nevertheless, Xu & Ding (2006) remark that wind forces can have significant effects

on the vehicle responses, especially in the lateral direction. Other studies of train

aerodynamics can be found in Chiu (1995) and Rangunathan et al. (2002). In Figure

4.52, the author shows a train exposed to a lateral cross wind pressure, while it travels

at a constant speed along the track.

Figure 4.52: Train exposed to a lateral cross wind pressure

Using the same three-dimensional vehicle properties of Section 4.4.5, the author

assumes that the train is travelling, along rigid rails, at a constant speed c of 45.55 m/s

(164 km/hr). The train is then exposed to a lateral cross wind pressure of 1.0 kN/m2

for a brief moment (0.95secs), before it dissipates to zero. In this example, time t is

arranged in such a manner that it takes 0.05 sec before the train is exposed to the cross

wind, then between the times 0.05 sec to 1.00 sec, the train experiences a positive

lateral pressure of 1.0 kN/m2, while beyond 1.00 sec, the lateral pressure returns to

zero. In Figure 4.53, a plot of the lateral displacement as a function of time for the 1st

and 7th

wheels of the train is plotted, while the lateral acceleration experienced by the

c

lateral wind pressure


193

vehicle body of the locomotive (V1) and first railway coach (V2) are shown in Figure

4.54. One can see a noticeable lateral shift as the pressure is applied followed by free

vibration when removed in Figures 4.53 and 4.54.

-0.10

-0.05

0.00

0.05

0.10

0.15

0.20

0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

Time (sec)

La

tera

l d

isp

lac

me

nt

(mm

)

3DTrain - 164 km/hr (W1)

3DTrain - 164 km/hr (W7)

Figure 4.53: Lateral displacement of the 1st and 7

th wheel of the train

Figure 4.54: Lateral acceleration experienced by the vehicle bodies of the train

Lateral pressure applied

Lateral pressure removed


194

4.5 Discussion of results and Conclusions

The main purpose of this chapter is the development and validation of the author’s

wheel-rail contact (WRC) elements. This system can be seen as an extension to

Chapter 3. Here the moving point force is replaced by a Hertzian spring. Additionally,

the author develops longitudinal and lateral springs to simulate the braking,

accelerating and lateral effects of passing trains. These additional features are current

unavailable for the ANSYS contact elements.

Examining the results presented in Section 4.3.1 to 4.3.5, the reader can see that the

WRC element and the ANSYS contact elements have similar results. In Section 4.3.2

and 4.3.5, it can be seen that both sprung wheel systems (WRC element and ANSYS

contact elements) can undergo a zero contact force, thus wheel-rail separation occurs;

nonetheless, as contact is re-established with the beam there is a large compressive

contact force experienced by the wheel. Section 4.3.6 then shows that a moving

sprung system can generate the same results as a moving unsprung system provided

that one uses reasonably large Hertzian spring stiffness between the wheel and rail,

especially at higher speed. One weakness of the ANSYS contact element method is

exposed in Section 4.3.7, which shows a loss of accuracy in the results as the number

of beam elements in the model decreases. However, this is not the case for the WRC

element, which maintain the correct solution. Additionally, Section 4.3.9 examines

the effects of braking on the bridge. This is also not possible using the ANSYS

contact elements. The results for the vertical response of the bridge and vehicle are in

good agreement with the results from the literature, while the results for the horizontal

response are also good and obey Newton’s second law of motion i.e. .F Ma=


195

Next using the developed WRC element, the author analyses the dynamic response of

the Boyne Viaduct railway bridge subjected to a single sprung wheel as well as to an

entire train consisting of a single six-axle locomotive and three four-axle railway

coaches. Each railway vehicle is modelled using lumped masses for the wheels,

elastic beam elements for the bogies and vehicle body, which are separated using

spring-dashpot elements, which were used to represent the primary and secondary

suspensions. The results for the single sprung wheel compare well with the results for

the moving load model, especially at low speeds 0.2α ≤ in the parametric study. At

realistic vehicle speeds the DAF results for the sprung wheel tend to be slightly larger

than the DAF results for the moving load due to its additional wheel inertia. This is

not the case for the two-dimensional and three-dimensional train models, which have

significantly lower dynamic amplification factors than the DAF for the multiple

moving loads. In all cases, the DAF experienced by the Boyne Viaduct never exceeds

1.1 at the range of speeds for this particular train type. Furthermore, the vertical

acceleration or riding comfort value of the railway coaches, travelling at a speed of

200 km/hr across the bridge, tend to lie below the recommended value of 1.0 m/s2

recommended in Eurocodes (1990).

Unlike the WRC element that uses rigid rails, the ANSYS contact element method

must use additional beam elements to the left and right of the bridge to support the

train as it enters and leaves the bridge structure, which means it requires additional

contact elements in the model. The braking and accelerating effects of a train as it

travels along the Boyne Viaduct, as presented in Section 4.4.4, are in good agreement

with the literature. Specifically, the reaction force increases significantly as each



196

drop in the force followed by a fluctuation of the results, until all wheels of the train

have exited the bridge and the reaction force oscillates about zero. For demonstration

purposes, the lateral spring element developed in Section 4.2.3 is used in Section 4.4.6

to provide lateral support to the wheels of the train, as the moving train is exposed to

lateral cross wind effects.

As a final note, it can be concluded that the WRC element is exceedingly robust in

capturing the dynamic response of both the bridge and vehicle. Additionally, one has

shown that the WRC element is comparable to, if not better than the commercial node-

to-surface contact elements in ANSYS. The WRC element can also model

irregularities. This is presented in Chapter 6.

Chapter 5 – Moving unsprung mass within a modal and finite element framework

197

Chapter 5

Moving unsprung mass represented by time varying

mass, damping, and stiffness matrices within a

modal and finite element framework

5.1 Introduction

As shown in the previous chapter, the sprung mass system is greatly influenced by the

Hertzian spring stiffness to the extent that large spring stiffness can cause the ANSYS

finite element program to become unstable during the transient analyses. However,

modelling the wheels of the train as moving unsprung masses can eliminate this

problem, where the Hertzian spring stiffness, kH, is infinite. In this chapter, the author

develops both a modal method as well as a finite element method for simulating a

moving unsprung mass traversing a bridge within the ANSYS framework.

When one deals with the moving unsprung mass its vertical position must be the same

as the vertical position of a point of the beam directly underneath. However, since the

moving unsprung mass is moving horizontally, its vertical velocity is not the same as

the vertical velocity of the point of the beam directly underneath, (which we will call


198

the local vertical velocity of the beam). In fact the vertical velocity of the moving

unsprung mass is equal to the local vertical velocity of the beam plus a convective

term. Furthermore, the vertical acceleration of the moving unsprung mass is equal to

the local vertical acceleration of the beam plus an additional convective term. Biggs

(1964) and Akin & Mofid (1989) present a moving unsprung model in which the

convective velocity and acceleration are omitted from the model; thus their solution is

inaccurate. This inaccuracy is overcome in the author’s study by including both the

local and convective terms for the moving unsprung mass.

This chapter considers the differential equation governing the vibration of a beam

subjected to the moving unsprung wheel and deriving a modal model with both the

local and convective acceleration of the unsprung mass taken into account. The use of

beam modes reduces the complexity of the solution. Next, the finite element method

for the moving unsprung mass is developed. This is different from the development of

the WRC element in Chapter 4. In this chapter additional mass, stiffness and damping

matrices are added to the beam matrix, whereas only additional stiffness matrices

were required in the WRC element. This chapter also highlights the similarities and

differences between the modal and finite element models from the point of view of

the form of the final matrices. The weighting and shape functions of the finite element

method play the same role as the mode shapes in the modal method. The difference

between the two methods occurs in the representation of the stiffness matrix.

Integration by parts gives the finite element stiffness matrix a different form from the

modal stiffness matrix. Additionally, the second derivatives of the beam element

shape functions are discontinuities in the finite element method.


199

In order to validate the modal and finite element models, the author compares results

with results from the literature as well as with results from use of the WRC element in

Chapter 4. In the WRC case, the Hertzian spring is given a reasonably large stiffness

and the moving sprung wheel is not allowed to lose contact with the beam i.e. no

separation. Additionally, this chapter highlights and quantifies the effects of omitting

the convective acceleration terms from an unsprung moving mass model as the results

obtained here are very different from Akin & Mofid’s (1989) results. Nevertheless,

many authors are still comparing the results of their models with this inaccurate

solution, ignoring the issue of convective acceleration (Bowe and Mullarkey, 2008).

Following the validation of both unsprung systems the author continues to investigate

the dynamic response to a single unsprung wheel and to multiple unsprung wheels

traversing the Boyne Viaduct and compares these responses with the responses to the

sprung wheel systems in Chapter 4 as well as with the responses to the moving loads

examined in Chapter 3.


200

5.2 Development of the unsprung wheel system

5.2.1 Modal superposition model incorporating a moving unsprung mass

The author begins by developing the modal superposition model for a moving

unsprung wheel Mw traversing a beam at a constant speed as presented in Figure 5.1.

Figure 5.1: Moving unsprung wheel traversing a beam

The differential equation governing the vibration of a beam subjected to the moving

unsprung wheel mass can be expressed as:

( ) ( )( )

4 2

4 2

, ,v x t v x tEI m p x

x t

∂ ∂+ =

∂ ∂ (5.1a)

whereby

( ) ( ) ( )( )p x F t x X tδ= − − (5.1b)

where ( )F t is the force imparted to moving unsprung mass by the beam, positive if

acting in the positive y-direction, the Dirac Delta function is represented byδ and the

position vector of the moving point mass is as follows:

( ) ( ) ( )t X t Y t= +p i j (5.2)

z, k

x, i

y, v, j

0

m, EI

( )dX tc

dt=

L

Mw


201

where i and j are the unit vector in the x and y direction, respectively. The horizontal

speed c of the moving mass, as shown in Figure 5.1, is:

( )dX tc

dt= (5.3)

The equation of motion for the moving mass in the y-direction is given as:

( )( )

2

2w w

d Y tM M g F t

dt= − + (5.4)

where g is the acceleration due to gravity. Eliminating ( )F t in Equation (5.1) using

(5.4) gives the following:

( ) ( ) ( )( )( )

4 2 2

4 2 2

, ,w w

v x t v x t d Y tEI m M M g x X t

x t dtδ

∂ ∂+ = − − −

∂ ∂ (5.5)

Since the moving mass is always in contact with the beam, the constraint equation is:

( ) ( )( )

,x X t

Y t v x t=

= (5.6)

Therefore the vertical velocity of the moving mass is as follows:

( ) ( )( )

( )( )

( ), ,x X t x X t

dY t v x t v x t dX t

dt t x dt= =

∂ ∂= +

∂ ∂ (5.7)

convective velocity


202

Furthermore the vertical acceleration of the moving mass is

( ) ( )( )

2 2

2 2

,x X t

d Y t v x t

dt t=

∂=

∂

( )( )

( ) ( )( )

( ) ( ) ( )( )

( )2 2 2

2 2

, , ,2

x X t x X t x X t

v x t dX t v x t dX t dX t v x t d X t

x t dt x dt dt x dt= = =

∂ ∂ ∂+ + +

∂ ∂ ∂ ∂ (5.8)

Substituting Equation (5.8) into (5.5) yields

( ) ( ) ( )( )

( )( )

( )4 2 2 2

4 2 2

, , , ,2

w x X t x X t

v x t v x t v x t v x t dX tEI m M

x t t x t dt= =

∂ ∂ ∂ ∂+ = − + +

∂ ∂ ∂ ∂ ∂

( )( )

( ) ( ) ( )( )

( )( )( )

2 2

2 2

, ,x X t x X t

v x t dX t dX t v x t d X tg x X t

x dt dt x dtδ

= =

∂ ∂+ + −

∂ ∂ (5.9)

In order to solve Equation (5.9), one uses the method of modal superposition;

whereby ( ),v x t can be represented according to Biggs (1964) as follows:

( ) ( ) ( )1

,N

n n

n

v x t r t xφ=

=∑ (5.10)

where ( )n xφ is the n-th characteristic shape as defined in Appendix C, and ( )nr t is

the n-th function of time which has to be calculated.

Substituting Equation (5.10) into (5.9) gives the following ordinary differential

equation governing ( )nr t , n = 1, 2, 3, … N.

convective acceleration


203

( ) ( ) ( ) ( ) ( ) ( )( )1 1 1

N N Niv

n n n n w n n

n n n

EI r t x m r t x M r t X tφ φ φ= = =

+ = −

∑ ∑ ∑&& &&

( )( ) ( )( )

( )( ) ( )( )

2

1 1

2N N

n n n n

n n

dX t dX tr t X t r t X t

dt dtφ φ

= =

′ ′′+ +

∑ ∑&

( )( ) ( )( ) ( )( )

2

21

N

n n

n

d X tr t X t g x X t

dtφ δ

=

′+ + −

∑ (5.11)

where a dot over ( )nr t represents a derivative with respect to time, and a dash over

( )n xφ represents a derivative with respect to x. Substituting Equation (C.7) and

(C.12b) into Equation (5.11) to eliminate the fourth derivative of ( )n xφ gives the

following:

( ) ( ) ( ) ( ) ( ) ( )( )4

1 1 1

N N N

n n n n n w n n

n n n

EI r t a x m r t x M r t X tφ φ φ= = =

+ = −

∑ ∑ ∑&& &&

( )( ) ( )( )

( )( ) ( )( )

2

1 1

2N N

n n n n

n n


dt dtφ φ

= =

′ ′′+ +

∑ ∑&

( )( ) ( )( ) ( )( )

2

21

N

n n

n


dtφ δ

=

′+ + −

∑ (5.12)

Multiplying both sides of Equation (5.12) by ( )i xφ , i = 1, 2, 3, … N, integrating

along the beam length, and using Equation (C.21) and (C.22) gives the following:

( ) ( ) ( ) ( )( )4

1 1 1

N N N

n n in n in w n n

n n n

EI r t a L m r t L M r t X tδ δ φ= = =

+ = −

∑ ∑ ∑&& &&

( )( ) ( )( )

( )( ) ( )( )

2

1 1

2N N

n n n n

n n


dt dtφ φ

= =

′ ′′+ +

∑ ∑&

( )( ) ( )( ) ( )( )

2

21

, = 1, 2, 3, ...N

n n i

n

d X tr t X t g X t i N

dtφ φ

=

′+ +

∑ (5.13)


204

where 1in

δ = when i n= , and 0in

δ = when i n≠ . Rearranging Equation (5.13) in the

order of the mass, damping and stiffness terms with the forcing term on the right-hand

side gives the following:

( )( ) ( )( ) ( )1

N

in w i n n

n

mL M X t X t r tδ φ φ=

+∑ &&

( )( )( ) ( )( ) ( )

1

2N

w i n n

n

dX tM X t X t r t

dtφ φ

=

′+

∑ &

( )( )( ) ( )( )

2

4

1

N

in n w i n

n

dX tEIL a M X t X t

dtδ φ φ

=

′′+ +

∑

( )( )( ) ( )( ) ( )

2

2w i n n

d X tM X t X t r t

dtφ φ

′+

( )( ), = 1, 2, 3, ...w iM g X t i Nφ= − (5.14)

Since Akin & Mofid (1989) only include the local accelerations of the moving

unsprung mass, terms containing ( )dX t dt and ( )2 2d X t dt are omitted from their

model. Expanding Equation (5.14) into matrix form gives:

1

2

3

1 0 0 ... 0

0 1 0 ... 0

0 0 1 ... 0

...... ... ... ... ...

0 0 0 ... 1 N

r

r

mL r

r

&&

&&

&&

&&

1 1 1 2 1 3 1 1

2 1 2 2 2 3 2 2

3 1 3 2 3 3 3 3

1 2 3

...

...

...

... ... ... ... ... ...

...

N

N

w N

N N N N N N

r

r

M r

r

φ φ φ φ φ φ φ φ




+

&&

&&

&&

&&

( )

1 1 1 2 1 3 1 1

2 1 2 2 2 3 2 2

3 1 3 2 3 3 3 3

1 2 3

...

...

2 ...

... ... ... ... ... ...

...

N

N

w N

N N N N N N

r

rdX t

M rdt

r





′ ′ ′ ′ ′ ′ ′ ′

′ ′ ′ ′ + ′ ′ ′ ′

&

&

&

&

411

422

433

4

0 0 ... 0

0 0 ... 0

0 0 ... 0

...... ... ... ... ...

0 0 0 ... NN

ra

ra

EIL ra

ra

+


205

( )

1 1 1 2 1 3 1 1

2 2 1 2 2 2 3 2 2

3 1 3 2 3 3 3 3

1 2 3

...

...

...

... ... ... ... ... ...

...

N

N

w N

N N N N N N

r

rdX t

M rdt

r





′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′

′′ ′′ ′′ ′′ + ′′ ′′ ′′ ′′

( )

1 1 1 2 1 3 1 1

2 1 2 2 2 3 2 22

3 1 3 2 3 3 3 32

1 2 3

...

...

...

... ... ... ... ... ...

...

N

N

w N

N N N N N N

r

rd X t

M rdt

r





′ ′ ′ ′ ′ ′ ′ ′

′ ′ ′ ′ + ′ ′ ′ ′

1

2

3

...

w

N

M g

φ

φ

φ

φ

= −

(5.15)

5.2.1.1 Modal superposition model of a moving unsprung load

Should one be only interested in the gravitational effects of a moving unsprung mass

i.e. a moving load, then the inertia effects of the moving unsprung mass can be

eliminated. This reduces Equation (5.15) to the following decoupling form:

1

2

3

1 0 0 ... 0

0 1 0 ... 0

0 0 1 ... 0

...... ... ... ... ...

0 0 0 ... 1 N

r

r

mL r

r

&&

&&

&&

&&

411

422

433

4

0 0 ... 0

0 0 ... 0

0 0 ... 0

...... ... ... ... ...

0 0 0 ... NN

ra

ra

EIL ra

ra

+

1

2

3

...

w

N

M g

φ

φ

φ

φ

= −

(5.16)

5.2.1.2 Unsprung wheel-rail separation

Despite being problematic, the author’s model can allow the unsprung wheel to

separate from the rail when the interaction between the wheel and the rail ceases to be

compressive, Equations (5.4) and (5.6) can be rewritten as:

( )( )2

2w w

d Y tF t M g M

dt= + (5.17a)

and


206

( ) ( )( )

,x X t

Y t v x t=

= (5.17b)

where ( )F t is the force imparted to the moving unsprung mass by the beam, positive

if acting in the positive y-direction. Equation (5.17b) is the constraint whereby the

unsprung wheel is in contact with the beam. The wheel is in contact with the rail

when ( )F t is positive, and separates from the rail when ( )F t becomes zero, when

the equation of motion of the unsprung wheel becomes:

( )2

20

w w

d Y tM g M

dt= + (5.17c)

At this stage, the constraint equation (5.17b) ceases to apply. It is replaced by the

separation distance d as follows:

( ) ( )( )

,x X t

d Y t v x t=

= − (5.17d)

During separation, Equation (5.17d) must be computed at every time-step in order to

track the vertical position of the unsprung wheel. The wheel remains separated from

the beam, while d is positive. When d becomes equal to zero, separation ceases, at

which time the constraint equation (5.17b) applies once again. Nevertheless, at the

instant the wheel regains contact with the beam; the beam is likely to experience a

large impact load from the wheel. In order to prevent this impact load, the author

would need to have a spring underneath the wheel, for a couple of time steps, so that

the wheel could regain contact with the beam at a gradual rate. However, this then

becomes a sprung wheel, which has been already developed in Chapter 4.


207

5.2.2 Finite element model incorporating the unsprung moving mass

The author now develops a finite element solution for the unsprung moving mass. In

Figure 5.2, the reader can see a free-body diagram of an unsprung wheel traversing a

beam element representing part of the bridge. The diagram indicates the coordinate

system where x is positive along the beam element, y is positive upward and z is

positive outwards. The origin of the coordinate system is at local node 1 of the beam.

For two-dimensional problems, the deflection in the y-direction and rotation about the

z-axis are defined by R1 and R2 at local node 1 of the beam, respectively, while R3 and

R4 are the deflection in the y-direction and rotation about the z-axis at local node 2 of

the beam. It should be noted that the author has intentionally redefined the nodal

displacement terms in this section so that the form of the finite element equations can

be compared with the form of the modal equations in Section 5.2.1.

Figure 5.2: Moving unsprung wheel traversing a beam element

Additionally, Equation (4.3) with a change of notation, computes the vertical

displacement at any point along the beam as follows:

( ) ( )4

1

, ( )n n

n

v x t R t x=

= Φ∑ (5.18)

R1 R3

R2

R4 z

x

y

0

m, EI

( )dX tc

dt=

Mw 1 2


208

where ( )1 1( )x N xΦ = , ( )2 1( )x G xΦ = , ( )3 2( )x N xΦ = , and ( )4 2( )x G xΦ = which are

defined by Equation (B.8) and plotted in Figure B.3 in Appendix B.

Substituting Equation (5.18) into (5.9) gives the following equation:

( ) ( ) ( ) ( )4 4

1 1

iv

n n n n

n n

EI R t x m R t x= =

Φ + Φ∑ ∑ &&

( ) ( )( )( )

( ) ( )( )4 4

1 1

2w n n n n

n n

dX tM R t X t R t X t

dt= =

′= − Φ + Φ

∑ ∑&& &

( )( ) ( )( )

( )( ) ( )( ) ( )( )

2 24 4

21 1

n nn n n

n n

dX t d X tR t X t R t X t g x X t

dt dtδ

= =

′′ ′+ Φ + Φ + −

∑ ∑ (5.19)

In order to apply Galerkin’s method of weighted residuals we multiply both sides of

Equation (5.19) by ( )i

xΦ , i = 1, 2, 3 and 4, and integrate along the element length;

( ) ( ) ( ) ( )4 4

1 10 0

( ) ( )

l l

iv

i n n i n n

n n

EI x x dxR t m x x dxR t= =

Φ Φ + Φ Φ =∑ ∑∫ ∫ &&

( )( ) ( )( )

( )( ) ( )4 4

1 10

( ) 2

l

w i n n n n

n n

dX tM x X t R t X t R t

dt= =

′= − Φ Φ + Φ

∑ ∑∫ && &

( )( )( ) ( )

( )( )( ) ( ) ( )( )

2 24 4

21 1

nn n n n

n n

dX t d X tX t R t X t R t g x X t dx

dt dtδ

= =

′′ ′+ Φ + Φ + −

∑ ∑ (5.20)

or

( ) ( )( ) ( )( ) ( )4

1 0

( )

l

i n w i n n

n

m x x dx M X t X t R t=

Φ Φ + Φ Φ +

∑ ∫ &&

( )( )( ) ( )( ) ( )

4

1

2w i n n

n

dX tM X t X t R t

dt=

′Φ Φ +

∑ &

( ) ( )( )

( )( ) ( )( )2

4

1 0

l

iv

i n w i nn

n

dX tEI x x dx M X t X t

dt=

′′Φ Φ + Φ Φ +

∑ ∫


209

( )( )( ) ( )( ) ( )

2

2w i n n

d X tM X t X t R t

dt

′Φ Φ

( )( ) , = 1, 2, 3, and 4w iM g X t i= − Φ (5.21)

The term ( ) ( )0

l

iv

i nx x dxΦ Φ∫ is now integrated by parts twice giving:

( ) ( ) ( ) ( ) ( ) ( )0

0 0

l lliv

i n i n i nx x dx x x x x dx′′′ ′ ′′′Φ Φ = Φ Φ − Φ Φ∫ ∫

( ) ( ) ( ) ( ) ( ) ( )0 0

0

ll l

i n i n i nx x x x x x dx′′′ ′ ′′ ′′ ′′= Φ Φ − Φ Φ + Φ Φ∫

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )0

00

l

i n i n i n i n i nl lx x x x x x x x x x dx′′′ ′′′ ′ ′′ ′ ′′ ′′ ′′= Φ Φ − Φ Φ − Φ Φ + Φ Φ + Φ Φ∫

(5.22)

Equation (5.22) implies the following:

( ) ( ) ( )4

1 0

l

iv

i n n

n

EI x x dxR t=

Φ Φ =∑ ∫

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )4

001

i n i n i n i nl ln

EI x x x x x x x x=

′′′ ′′′ ′ ′′ ′ ′′Φ Φ − Φ Φ − Φ Φ + Φ Φ∑

( ) ( ) ( )0

l

i n nx x dx R t′′ ′′+ Φ Φ∫ (5.23)

By means of Equation (A.16) and (A.19), Equation (5.23) becomes:

( ) ( ) ( )4

1 0

l

iv

i n n

n

EI x x dxR t=

Φ Φ =∑ ∫

( ) ( ) ( ) ( )00

i i i il lx Q x Q x M x M′ ′−Φ + Φ − Φ + Φ ( ) ( ) ( )

4

1 0

l

i n n

n

EI x x dxR t=

′′ ′′+ Φ Φ∑ ∫ (5.24)

Substituting Equation (B.1) into (5.24) gives:

( ) ( ) ( )4

1 0

l

iv

i n n

n

EI x x dxR t=

Φ Φ =∑ ∫


210

( ) ( ) ( ) ( ) ( ) ( ) ( )4

2 1 2 1

1 0

ˆ ˆ ˆ ˆ0 0

l

i i i i i n n

n

l Q Q l M M EI x x dxR t=

′ ′ ′′ ′′−Φ − Φ − Φ − Φ + Φ Φ∑ ∫ (5.25)

Substituting Equation (5.25) into (5.21) gives the following equation as:

( ) ( )( ) ( )( ) ( )4

1 0

( )

l

i n w i n n

n


Φ Φ + Φ Φ +

∑ ∫ &&

( )( )( ) ( )( ) ( )

4

1

2w i n n

n

dX tM X t X t R t

dt=

′Φ Φ +

∑ &

( ) ( )( )

( )( ) ( )( )2

4

1 0

l

i n w i nn

n


dt=

′′ ′′ ′′Φ Φ + Φ Φ +

∑ ∫

( )( )( ) ( )( ) ( )

2

2w i n n

d X tM X t X t R t

dt

′Φ Φ

( )( ) ( ) ( ) ( ) ( )1 2 1 2ˆ ˆ ˆ ˆ0 0 , = 1, 2, 3, and 4

w i i i i iM g X t Q l Q M l M i′ ′= − Φ + Φ + Φ + Φ + Φ (5.26)

Expanding Equation (5.26) into matrix form gives:

1 1 1 2 1 3 1 4 1 1 1 2 1 3 1 41 1

2 1 2 2 2 3 2 4 2 1 2 2 2 3 2 42 2

3 1 3 2 3 3 3 4 3 1 3 2 3 3 3 430

4 1 4 2 4 3 4 4 4 1 4 2 4 3 4 44

l

w

R R

R Rm dx M

R

R

Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ +

Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ

∫

&& &&

&& &&

&& &

&&

3

4

R

R

&

&&

( )1 1 1 2 1 3 1 4 1 1 1 2 1 3 1 41

2 1 2 2 2 3 2 4 2 1 2 2 2 3 2 42

3 1 3 2 3 3 3 4 33

4 1 4 2 4 3 4 4 4

2w

R

dX t RM EI

dt R

R

′ ′ ′ ′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ ′ ′ ′ ′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ + +

′ ′ ′ ′ ′′ ′′ Φ Φ Φ Φ Φ Φ Φ Φ Φ ′ ′ ′ ′Φ Φ Φ Φ Φ Φ Φ Φ

&

&

&

&

1

2

1 3 2 3 3 3 4 30

4 1 4 2 4 3 4 4 4

l

R

Rdx

R

R

′′ ′′ ′′ ′′ ′′ ′′ Φ Φ Φ Φ Φ Φ Φ

′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′Φ Φ Φ Φ Φ Φ Φ Φ

∫

( ) ( )1 1 1 2 1 3 1 4 1 1 1 2 1 3 1 41

2 2

2 1 2 2 2 3 2 4 2 1 2 2 2 3 2 42

2

3 1 3 2 3 3 3 4 3 13

4 1 4 2 4 3 4 4 4

w w

R

RdX t d X tM M

Rdt dt

R

′′ ′′ ′′ ′′ ′ ′ ′ ′ΦΦ ΦΦ ΦΦ ΦΦ ΦΦ ΦΦ ΦΦ ΦΦ ′′ ′′ ′′ ′′ ′ ′ ′ ′Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ + +

′′ ′′ ′′ ′′ ′ Φ Φ Φ Φ Φ Φ Φ Φ Φ Φ ′′ ′′ ′′ ′′Φ Φ Φ Φ Φ Φ Φ Φ

1

2

3 2 3 3 3 4 3

4 1 4 2 4 3 4 4 4

R

R

R

R

′ ′ ′ Φ Φ Φ Φ Φ Φ

′ ′ ′ ′Φ Φ Φ Φ Φ Φ Φ Φ

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

1 1 1 11

2 2 2 22

1 2 1 2

3 3 3 33

4 4 4 44

0 0

0 0ˆ ˆ ˆ ˆ

0 0

0 0

w

l l

l lM g Q Q M M

l l

l l

′ ′Φ Φ Φ Φ Φ ′ ′Φ Φ Φ ΦΦ

= − + + + + ′ ′Φ Φ Φ ΦΦ

′ ′Φ Φ Φ ΦΦ

(5.27)


211

With the aid of Figure B.3, Equation (5.27) reduces to the following:

1 1 1 2 1 3 1 4 1 1 1 2 1 3 1 41 1

2 1 2 2 2 3 2 4 2 1 2 2 2 3 2 42 2

3 1 3 2 3 3 3 4 3 1 3 2 3 3 3 430

4 1 4 2 4 3 4 4 4 1 4 2 4 3 4 44

l

w

R R

R Rm dx M

R

R



∫

&& &&

&& &&

&& &

&&

3

4

R

R

&

&&

( )1 1 1 2 1 3 1 4 1 1 1 2 1 3 1 41

2 1 2 2 2 3 2 4 2 1 2 2 2 3 2 42

3 1 3 2 3 3 3 4 33

4 1 4 2 4 3 4 4 4

2w

R

dX t RM EI

dt R

R



&

&

&

&

1

2

1 3 2 3 3 3 4 30

4 1 4 2 4 3 4 4 4

l

R

Rdx

R

R

′′ ′′ ′′ ′′ ′′ ′′ Φ Φ Φ Φ Φ Φ Φ

′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′Φ Φ Φ Φ Φ Φ Φ Φ

∫

( ) ( )1 1 1 2 1 3 1 4 1 1 1 2 1 3 1 41

2 2

2 1 2 2 2 3 2 4 2 1 2 2 2 3 2 42

2

3 1 3 2 3 3 3 4 3 13

4 1 4 2 4 3 4 4 4

w w

R

RdX t d X tM M

Rdt dt

R



1

2

3 2 3 3 3 4 3

4 1 4 2 4 3 4 4 4

R

R

R

R

′ ′ ′ Φ Φ Φ Φ Φ Φ

′ ′ ′ ′Φ Φ Φ Φ Φ Φ Φ Φ

1

2

1 2 1 2

3

4

1 0 0 0

0 0 1 0ˆ ˆ ˆ ˆ0 1 0 0

0 0 0 1

wM g Q Q M M

Φ Φ

= − + + + + Φ Φ

(5.28)

Comparing Equation (5.15) with (5.28), one can clearly see that the modal equation is

remarkably similar to the finite element equation apart from the representation of the

stiffness matrix and the nodal forces, 1 2 1 2ˆ ˆ ˆ ˆ, , and Q Q M M ; moreover, during the finite

element assembly Newton’s Third Law eliminates these forces at internal nodes

(Bowe & Mullarkey, 2008). In addition, the weighting and shape functions of the

finite element model play a similar role as the mode shapes play in the analytical

numerical solution. The difference between the two models occurs in the

representation of the stiffness matrix. Integration by parts gives the finite element

stiffness matrix a different form to the modal stiffness matrix. Furthermore, the

second derivatives of the beam element shape functions are discontinuous in the finite

element model.


212

5.2.3 Modal superposition for multiple unsprung masses traversing a beam

In this sub-section, the author expands the single moving unsprung mass to several

moving masses traversing a beam. Recalling Equation (5.11), the equation of motion

for a single moving unsprung mass crossing a beam can be rewritten as follows:

( ) ( ) ( ) ( ) ( ) ( )( )1 1 1

N N Niv

n n n n w n n

n n n

EI r t x m r t x M r t X tφ φ φ= = =

+ = −

∑ ∑ ∑&& &&

( )( ) ( )( )

( )( ) ( )( )

2

1 1

2N N

n n n n

n n


dt dtφ φ

= =

′ ′′+ +

∑ ∑&

( )( ) ( )( ) ( )( )

2

21

N

n n

n


dtφ δ

=

′+ + −

∑ (5.29)

For several wheels, ( )X t will denote the horizontal position of the front wheel of

train; hence, the position of the m-th wheel, of the k-th bogie, of the j-th carriage of

the train and is equal to ( ) ( ) ( ) ( )1 1 1 ,w w w

X t D j B k A m− − − − − − where ( ) .X t ct=

Updating Equation (5.29) for several moving unsprung masses then gives:

( ) ( ) ( ) ( )1 1

N Niv

n n n n

n n

EI r t x m r t xφ φ= =

+ =∑ ∑ &&

( ) ( ) ( ) ( ) ( ) ( )( )2 2

, ,1 1 1 1

1 1 1cN N

n n w w ww j k m

j k m n

M r t X t D j B k A mφ= = = =

− − − − − − −

∑∑∑ ∑ &&

( )( ) ( ) ( ) ( ) ( )( )

1

2 1 1 1N

n n w w w

n

dX tr t X t D j B k A m

dtφ

=

′+ − − − − − −∑ &

( )( ) ( ) ( ) ( ) ( )( )

2

1

1 1 1N

n n w w w

n


dtφ

=

′′+ − − − − − −

∑

( )( ) ( ) ( ) ( ) ( )( )

2

21

1 1 1N

n n w w w

n

d X tr t X t D j B k A m

dtφ

=

′+ − − − − − −∑

] ( ) ( ) ( ) ( )( )1 1 1w w wg x X t D j B k A mδ+ − + − + − + − (5.30)


213

where ( ), ,w j k mM is the unsprung mass of the m-th wheel, of the k-th bogie, of the j-th

carriage of the train. With the aid of Equations (3.40a) and (3.40b), the m-th wheel, of

the k-th bogie, of the j-th carriage of the train is only on the beam between the times

( ) ( ) ( )1 1 1w w w

D j B k A m

c

− + − + − to

( ) ( ) ( )1 1 1.

w w wL D j B k A m

c

+ − + − + − Following

a similar methodology to Section 5.2.1, Equations (C.7) and (C.12b) are substituted

into Equation (5.30) to eliminate the fourth derivative of ( )n xφ giving:

( ) ( ) ( ) ( )4

1 1

N N

n n n n n

n n

EI r t a x m r t xφ φ= =

+ =∑ ∑ &&

( ) ( ) ( ) ( ) ( ) ( )( )2 2

, ,1 1 1 1

1 1 1cN N

n n w w ww j k m

j k m n


− − − − − − −

∑∑∑ ∑ &&

( )( ) ( ) ( ) ( ) ( )( )

1

2 1 1 1N

n n w w w

n


dtφ

=

′+ − − − − − −∑ &

( )( ) ( ) ( ) ( ) ( )( )

2

1

1 1 1N

n n w w w

n


dtφ

=

′′+ − − − − − −

∑

( )( ) ( ) ( ) ( ) ( )( )

2

21

1 1 1N

n n w w w

n


dtφ

=

′+ − − − − − −∑

] ( ) ( ) ( ) ( )( )1 1 1w w wg x X t D j B k A mδ+ − + − + − + − (5.31)

Multiplying both sides of Equation (5.31) by ( )i xφ , i = 1, 2, 3, … N, integrating

along the beam length, and using Equation (C.21) and (C.22) gives the following:

( ) ( )4

1 1

N N

n n in n in

n n

EI r t a L m r t Lδ δ= =

+ =∑ ∑ &&

( ) ( ) ( ) ( ) ( ) ( )( )2 2

, ,1 1 1 1

1 1 1cN N

n n w w ww j k m

j k m n


− − − − − − −

∑∑∑ ∑ &&

( )( ) ( ) ( ) ( ) ( )( )

1

2 1 1 1N

n n w w w

n


dtφ

=

′+ − − − − − −∑ &


214

( )( ) ( ) ( ) ( ) ( )( )

2

1

1 1 1N

n n w w w

n


dtφ

=

′′+ − − − − − −

∑

( )( ) ( ) ( ) ( ) ( )( )

2

21

1 1 1N

n n w w w

n


dtφ

=

′+ − − − − − −∑

] ( ) ( ) ( ) ( )( )1 1 1 , = 1, 2, 3, ...i w w wg X t D j B k A m i Nφ+ − − − − − − (5.32)

where 1in





( ) ( ) ( ) ( ) ( )( )2 2

, ,1 1 1 1

1 1 1cNN

in i w w ww j k mn j k m

mL M X t D j B k A mδ φ= = = =

+ − − − − − −

∑ ∑∑∑

( ) ( ) ( ) ( )( ) ( )1 1 1n w w w n

X t D j B k A m r tφ⋅ − − − − − − &&

( )( ) ( ) ( ) ( ) ( )( )

2 2

, ,1 1 1 1

2 1 1 1cNN

i w w ww j k m

n j k m

dX tM X t D j B k A m

dtφ

= = = =

+ − − − − − −

∑ ∑∑∑

( ) ( ) ( ) ( )( ) ( )1 1 1n w w w n

X t D j B k A m r tφ′⋅ − − − − − − &

( )

( )2

2 24

, ,1 1 1 1

cNN

in n w j k mn j k m

dX tEIL a M

dtδ

= = = =

+ +

∑ ∑∑∑

( ) ( ) ( ) ( )( ) 1 1 1i w w wX t D j B k A mφ⋅ − − − − − −

( ) ( ) ( ) ( )( ) 1 1 1n w w wX t D j B k A mφ′′⋅ − − − − − −

( )

( )( ) ( ) ( ) ( )( )

22 2

, , 21 1 1

1 1 1cN

i w w ww j k mj k m

d X tM X t D j B k A m

dtφ

= = =

+ − − − − − −∑∑∑

( ) ( ) ( ) ( )( ) ( ) 1 1 1n w w w nX t D j B k A m r tφ′⋅ − − − − − −

( ) ( ) ( ) ( ) ( )( )2 2

, ,1 1 1

1 1 1 , = 1, 2, 3, ...cN

i w w ww j k mj k m

M g X t D j B k A m i Nφ= = =

= − − − − − − −∑∑∑ (5.33)



215

1

2

3

1 0 0 ... 0

0 1 0 ... 0

0 0 1 ... 0

...... ... ... ... ...

0 0 0 ... 1 N

r

r

mL r

r

&&

&&

&&

&&

( )

1 1 1 2 1 3 1 1

2 1 2 2 2 3 2 22 2

3 1 3 2 3 3 3 3, ,1 1 1

1 2 3

...

...

...

... ... ... ... ... ...

...

c

N

NN

Nw j k mj k m

N N N N N N

r

r

M r

r





= = =

+

∑∑∑

&&

&&

&&

&&

( )

( )

1 1 1 2 1 3 1 1

2 1 2 2 2 3 2 22 2

3 1 3 2 3 3 3 3, ,1 1 1

1 2 3

...

...

2 ...

... ... ... ... ... ...

...

c

N

NN

Nw j k mj k m

N N N N N N

r

rdX t

M rdt

r





= = =

′ ′ ′ ′ ′ ′ ′ ′

′ ′ ′ ′ + ′ ′ ′ ′

∑∑∑

&

&

&

&

411

422

433

4

0 0 ... 0

0 0 ... 0

0 0 ... 0

...... ... ... ... ...

0 0 0 ... NN

ra

ra

EIL ra

ra

+

( )

( )

1 1 1 2 1 3 1 1

2 2 1 2 2 2 3 2 22 2

3 1 3 2 3 3 3 3, ,1 1 1

1 2 3

...

...

...

... ... ... ... ... ...

...

c

N

NN

Nw j k mj k m

N N N N N N

r

rdX t

M rdt

r





= = =

′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′

′′ ′′ ′′ ′′ + ′′ ′′ ′′ ′′

∑∑∑

( )

( )

1 1 1 2 1 3 1 1

2 1 2 2 2 3 2 222 2

3 1 3 2 3 3 3 3, , 21 1 1

1 2 3

...

...

...

... ... ... ... ... ...

...

c

N

NN

Nw j k mj k m

N N N N N N

r

rd X t

M rdt

r





= = =

′ ′ ′ ′ ′ ′ ′ ′

′ ′ ′ ′ + ′ ′ ′ ′

∑∑∑

( )

1

22 2

3, ,1 1 1

...

cN

w j k mj k m

N

M g

φ

φ

φ

φ

= = =

= −

∑∑∑ (5.34)

In the matrices, 1 2 3, , φ φ φ etc. are evaluated at ( ) ( ) ( ) ( )1 1 1 .w w w

X t D j B k A m− − − − − −


216

5.3 Validating the Unsprung Mass Systems

In order to validate modal and finite element unsprung mass models, the author

compares their results with suitable results obtained from the literature. In all

examples, the gravitational and damping effects of the beam are ignored. In addition,

one uses the Newmark time integration method with 500 equal time steps to solve the

transient analysis and time t is arranged in such a manner that the front wheel of the

train is at the left support at t = 0 sec. The initial displacement and velocity of the

unsprung wheel and beam are equal to zero. It should be noted that in each example,

unless otherwise stated, the author uses 6 modes in the modal system, while in the

finite element system the beam is discretized into 10 beam elements.

5.3.1 Wheel as a moving unsprung load traversing a cantilever beam

In this first example, the author examines the effects of a moving unsprung mass

traversing a fixed-free cantilever beam at constant speed c of 50.8 m/s from left to

right, where the inertia effects of the wheel mass are omitted; thus only the

gravitational effects of the mass are taken into account i.e. the mass of the wheel Mw

is equal to 2629 kg, thus the wheel has a weight of 25.79 kN. The author adopts the

same cantilever beam as Akin & Mofid (1989), such that the beam has a length L of

7.62 m, Young modulus of elasticity E of 2.07 x 1011

N/m2, moment of inertia I of

4.58 x 10-5

m4, and mass per unit length m of 46 kg/m, so .7.5

wM mL = Using

Equation (C.27a) the first natural frequency of this particular beam is

1 27.49 rad/ c.seω = It should also be noted that the beam properties used are very

responsive and unrealistic in practise and are only used for comparison purposes.

The vertical displacement at the free-end of the cantilever beam is plotted as a

function of time in Figure 5.5. The results shown in Figure 5.5a relate to the moving


217

unsprung load traversing a fixed-free cantilever beam (fixed at x = 0), while Figure

5.5b shows the results of the moving unsprung load traversing a free-fixed cantilever

(fixed at x = L). One can observe a striking similarity between the results from both

for the author’s system and the results obtained from Akin & Mofid (1989). The

reader should be aware that in Akin & Mofid’s (1989) paper, gravity acts in the

upwards direction. In addition, one can also conclude that all results are identical to

those obtained in Figure 3.12 for the exact and simple model.

-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

Akin & Mofid (1989)

Modal (6 Modes)

Finite Element Model

-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

Akin & Mofid (1989)

Modal (6 Modes)


Figure 5.5: Vertical displacement at the free-end of the cantilever beam due to a

moving unsprung load: (a) Fixed-free cantilever beam; (b) Free-fixed

cantilever beam (Bowe & Mullarkey, 2008)

(a)

(b)


218

5.3.2 Wheel as a moving unsprung mass traversing a cantilever beam

Using the same beam and point mass properties as those of Section 5.3.1, one now

investigates the effects of a moving unsprung mass traversing a cantilever beam as

shown in Figure 5.6. On this occasion the inertia effects of the wheel mass as well as

its gravitational effects are taken into account. The wheel is traversing at constant

speed c = 50.8 m/s from left to right across the beam. In this particular model, the

unsprung wheel has a mass Mw = 2629 kg, i.e. 7.5.w

M mL =

Figure 5.6: Moving unsprung mass traversing a cantilever beam

In Figure 5.7, the author plots the vertical displacement at the free-end of the fixed-

free cantilever beam (fixed at x = 0) as a function of time. Only the local acceleration

of the moving unsprung wheel is taken into account in Figure 5.7a, while Figure 5.7b

presents the results with the total (local plus convective) acceleration of the moving

unsprung wheel. In addition, the author plots in Figure 5.7b the results of a moving

sprung wheel traversing the cantilever, as developed in Chapter 4, where the Hertzian

spring is given a reasonably large stiffness value of 1.4x106 kN/m. In one example,

the moving sprung wheel is not allowed to separate from the beam; in another, it is

allowed to separate. Examining the results of Figure 5.7a, it can be seen that the

results of the author’s model are quite similar to the results of Akin & Mofid’s (1989)

modal system when the author omits convective acceleration. The results for the

author’s two unsprung models in Figure 5.7b show a striking similarity with the

z

y

0

L

Mw

x

c


219

results from the sprung model with no separation. Figure 5.7b shows that the results

from the sprung model with separation are slightly different.

-0.30

-0.25

-0.20

-0.15

-0.10

-0.05

0.00

0.05

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

Akin & Mofid (1989)

Modal (6 Modes) - Local acceleration only

Finite Element Model - Local acceleration only

-0.07

-0.06

-0.05

-0.04

-0.03

-0.02

-0.01

0.00

0.01

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t m

)

Modal (6 Modes) - Total acceleration

Finite Element Model - Total acceleration

Sprung Wheel - Without seperation

Sprung Wheel - With seperation (Figure 4.8a)

Figure 5.7: Vertical displacement of the free end of a fixed-free cantilever beam due to a

moving unsprung wheel travelling at 50.8 m/s: (a) local acceleration only;

(b) total (local plus convective) acceleration (Bowe & Mullarkey, 2008)

The reader can see by comparing Figure 5.7a with 5.7b that convective acceleration

should not be omitted. It can be seen that the free-end deflection in Figure 5.7b is a

quarter the deflection in Figure 5.7a at time t = 0.15 sec. Moreover, the graphs in

Figure 5.7a are different in form from the graphs of Figure 5.7b.

(a)

(b)


220

In Figure 5.8, one can see that the contact force between the moving unsprung wheel

and the beam goes negative at time t = 0.085 sec. In the case of the moving sprung

wheel, separation occurs at this time; contact is re-established at time t = 0.114 sec.

For the moving unsprung systems, it is shown that the reaction force becomes

negative at a much earlier time t = 0.099 sec. During separation, the reaction force of

the unsprung wheel must still be computed in order to track its vertical position.

-2.0

-1.5

-1.0

-0.5

0.0

0.5

0 0.03 0.06 0.09 0.12 0.15

Time (sec)

Co

nta

ct

Fo

rce

/ S

tati

c W

eig

ht



Sprung Wheel - Without Seperation

Sprung Wheel - With Seperation (Figure 4.9a)

Figure 5.8: Contact force between the moving unsprung mass and beam

Next the authors examine the dynamic effects of the moving unsprung wheel

traversing a free-fixed cantilever beam (fixed at x = L) as shown in Figure 5.9.

Figure 5.9: Moving unsprung mass traversing a free-fixed cantilever beam

z

y

0

L

Mw

x

c


221

The vertical displacement at the free-end of the cantilever beam as a function of time

is plotted in Figure 5.10. As before, only the local acceleration of the moving

unsprung wheel is taken into account in Figure 5.19a, while Figure 5.10b presents

results for the total (local plus convective) acceleration of the moving unsprung wheel.

-0.12

-0.10

-0.08

-0.06

-0.04

-0.02

0.00

0.02

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

Akin & Mofid (1989)

Modal (6 Modes) - Local acceleration only

Finite Element Model - Local acceleration only

-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.00 0.03 0.06 0.09 0.12 0.15

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)




Sprung Wheel - With seperation (Figure 4.8b)

Figure 5.10: Vertical displacement of the free-end of a free-fixed cantilever beam due to

a moving unsprung mass travelling at 50.8 m/s: (a) local acceleration only;

(b) total (local plus convective) acceleration (Bowe & Mullarkey, 2008)

Examining the results of Figure 5.10a, it can be clearly seen that the results of the

author’s model are very similar to the results of Akin & Mofid’s (1989) modal

(a)

(b)


222

system, while the results for the author’s unsprung model in Figure 5.10b are

comparable with the results from the sprung model with no separation. Comparing the

free-end deflection of the beam in Figure 5.10a with 5.10b, it can be seen that the

results vary by a factor of 4, which is a similar ratio to that observed in Figure 5.7.

It can be observed from Figure 5.11 that as the moving unsprung wheel approaches

the right-hand support of the cantilever beam, it experiences a substantial compressive

reaction force from the beam until time t = 0.135 sec. Beyond time t = 0.135 sec the

wheel then experiences a very large tensile reaction force from the beam. In the case

of the moving sprung wheel, separation occurs at this time and contact with the beam

is not re-established for the remaining duration of this analysis.

-20

-15

-10

-5

0

5

10

15

20

0 0.03 0.06 0.09 0.12 0.15

Time (sec)

Co

nta

ct

Fo

rce

/ S

tati

c W

eig

ht

Modal (6 Modes) - Total Acceleration

Finite Element Model - Total Acceleration


Sprung Wheel - With seperation (Figure 4.9b)

Figure 5.11: Contact force between the moving unsprung mass and beam

The author now examine the effects of convective accleration at lower speeds. Figure

5.12 presents the vertical displacement at the free-end of a cantilever beam as a

function of time where the cantilever is subjected to an unsprung mass traversing the

cantilever at a speed c of 27.78 m/s as well as at a very slow speed c of 1 m/s using


223

the author’s models. The results obtained from the fixed-free cantilever beam (fixed at

x = 0) are shown in Figure 5.12a, while Figures 5.12b and 5.12c plot the results from

the free-fixed cantilever beam (fixed at x = L) due to the moving unsprung mass.

Examining Figures 5.12a and 5.12b, the reader can see that the free-end deflection of

the cantilever beam without convective acceleration varies by a factor of 3 with the

results for the total (local plus convective) acceleration at the speed c of 27.78 m/s.

Whereas, at the very slow speed c of 1 m/s, Figure 5.12a shows that there are no

significant difference between the convective acceleration and the total (local plus

convective) acceleration, while Figure 5.12c shows a noticeable difference in results

at lower speeds.

It can be seen from Figure 5.12c that as the moving unsprung mass enters the free-end

of the free-fixed cantilever beam it tends to undergo a periodic oscillation; however,

the local acceleration model tends to decay at a much quicker rate than the total (local

plus convective) acceleration model as the moving unsprung mass approaches the

support (Bowe & Mullarkey, 2008).

-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0


Ve

rtic

al D

isp

lac

em

en

t (m

)

Modal & FEM - Local acceleration (1m/s)

Modal & FEM - Local acceleration (27.78 m/s)

Modal & FEM - Local plus convective acceleration (1m/s)

Modal & FEM - Local plus convective acceleration (27.78 m/s)

(a)


224

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.1

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0


Ve

rtic

al D

isp

lac

em

en

t (m

)

Modal & FEM - Local acceleration (27.78 m/s)

Modal & FEM - Local plus convective acceleration (27.78 m/s)

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0


Ve

rtic

al D

isp

lac

em

en

t (m

)

Modal & FEM - Local acceleration (1 m/s)

Modal & FEM - Local plus convective acceleration (1 m/s)

Figure 5.12: Vertical displacement at the free-end of a cantilever beam due to a

moving unsprung mass at various speeds: (a) fixed-free cantilever

beam at both speeds; (b) free-fixed cantilever beam at 27.78 m/s; (c)

free-fixed cantilever beam at 1 m/s (Bowe & Mullarkey, 2008)

5.3.3 Moving unsprung system traversing a simply supported beam

For comparison purposes, one now examines a simply supported beam subjected to a

single moving system as presented in Figure 5.13. This system comprises two masses,


225

one an unsprung mass Mw in contact with beam and the other a sprung mass Mv

supported by means of a spring of stiffness k1. In accordance with Fryba (1999), the

author will now show that varying the ratio between the sprung mass Mv and the

unsprung mass Mw will have little dynamic effect on the bridge. In all simulations, the

system traverses the beam at a constant speed c of 27.78 m/s (100 km/hr). The total

mass of the vehicle ( )w vM M+ is equal to 5.75t, while the spring stiffness k1 is 1595

kN/m. The different w v

M M ratios examined by the author are: 0.01, 0.25, 1.0, 4.0

and 100. Using the same data as Yang and Wu (2001), the beam has a length L of

25m, Young’s modulus of elasticity E of 2.87x106 kN/m

2, moment of inertia I of 2.9

m4, mass per unit length m of 2.303 t/m and a Poisson’s ratio ν of 0.2. As in previous

models, the gravitational and damping effects of the beam are ignored. In each case,

time t is arranged in such a manner that the wheel is at the left hand support at t = 0

sec and the initial displacement and velocity of the beam are equal to zero. In

addition, the author describes the mid-span deflection of the simply supported beam

using the first 6 modes of vibration in the modal system, while in the finite element

system the beam is discretized into 10 beam elements. The Newmark time integration

method (Bathe, 1996) with 500 equal time steps is used to solve the transient analysis.

In this particular model, the vehicle to beam mass ratio is ( ) 0.01.w vM M mL+ =

Figure 5.13: Moving unsprung system traversing a simply supported beam

z

y

0

L

x

c

Mw

Mv

k1


226

In Figure 5.14, the author presents the vertical displacement, vertical acceleration and

bending moment at mid-point of the beam as a function of time for both the modal

and finite element models. Since the w v

M M ratio is relatively small, comparisons

with the results can also be made with the results presented in Section 4.3.4 for the

sprung mass model. For the internal forces in finite element models, the author has

only chosen local node 2 of the 5th

element (see Section 3.2.2.3 for details about

internal forces). From inspection of Figure 5.14, one can see that all results are very

similar with each other.

-0.0030

-0.0025

-0.0020

-0.0015

-0.0010

-0.0005

0.0000

0.0005

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

Modal (6 modes)


Sprung Wheel - Figure 4.14a

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

0.4

0.5

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al A

cc

ele

rati

on

(m

/s2)

Modal (6 modes) Finite Element Model Sprung Wheel - Figure 4.14b

(a)

(b)


227

0

50

100

150

200

250

300

350

400

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Be

nd

ing

Mo

me

nt

(kN

m)

Modal (6 modes)


Sprung Wheel - Figure 4.14c

Figure 5.14: Time history at mid-span of the beam: (a) vertical displacement; (b)

vertical acceleration; (c) bending moment

The vertical displacements at mid-span of the beam as a function of time, for the

remaining w v

M M ratios, are then plotted in Figure 5.15. Examining the results in

Figure 5.15, one can see that the results for the modal method are very similar with

the results for the finite element models. In addition, one can see that varying the ratio

between the sprung and unsprung masses tends to have little effect on the dynamic

response of the beam, as observed by Fryba (1999), for these specific values.

-0.0030

-0.0025

-0.0020

-0.0015

-0.0010

-0.0005

0.0000

0.0005

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

Modal (6 modes) - Mw /Mv=0.25

Finite Element Model - Mw /Mv=0.25

(c)

(a)


228

-0.0030

-0.0025

-0.0020

-0.0015

-0.0010

-0.0005

0.0000

0.0005

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)



-0.0030

-0.0025

-0.0020

-0.0015

-0.0010

-0.0005

0.0000

0.0005

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)



-0.0030

-0.0025

-0.0020

-0.0015

-0.0010

-0.0005

0.0000

0.0005

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

Modal (6 modes) - Mw /Mv=100

Finite Element Model - Mw /Mv=100

Figure 5.15: Vertical displacement at mid-span of simply supported beam as a

function of time due to a moving unsprung system with various

w vM M ratios: (a) 0.25; (b) 1.00; (c) 4.00; (d) 100

(b)

(c)

(d)


229

5.3.4 Unsprung wheel system at wide range of speeds

Adopting the same simply supported beam properties to that of Yang & Wu (2001),

such that bridge has a length L of 25m, Young’s modulus of elasticity E of 2.87x106

kN/m2, moment of inertia I of 2.9 m

4, mass per unit length m of 2.303 t/m and a

Poisson’s ratio ν of 0.2, the author now considers the following unsprung vehicle

model as shown in Figure 5.16. The vehicle to bridge mass ratio is 0.5; the unsprung

wheel mass to sprung vehicle mass ratio is 0.25; the bridge to vehicle frequency ratio

is 3; the vehicle damping ratio is 0.125. These values are the same as those used by

Olsson (1985); thus, in this example Mw is 5.75 t, Mv is 23 t, k1 is 2300 kN/m, and c1

is 57.5 kNs/m. In Figure 5.17, the author plots the dynamic amplification factor at

mid-span of the beam for a wide range of speeds for the moving unsprung vehicle.

From the results, one can observe that the results of the modal and finite element

system are remarkably similar to Olsson’s (1985) results. Comparing Figure 5.17 with

Figure 4.22, the author can reinforce the point that the unsprung model is comparable

to the sprung model provided that the Hertzian spring stiffness is given a reasonably

large value.

Figure 5.16: Moving unsprung vehicle traversing a simply supported beam

z

y

0

L

x

c

Mw

Mv

k1 c1


230

1.0

1.1

1.2

1.3

1.4

1.5

1.6

1.7

0.0 0.2 0.4 0.6 0.8 1.0

Velocity Ratio αααα

Dy

na

mic

am

plifi

ca

tio

n f

ac

tor D

d

Olsson [10] Unsprung System

Modal (6 Modes)


Figure 5.17: Moving unsprung vehicle at a wide range of speeds

5.3.5 Sensitivity analysis of the unsprung systems and simply supported beam

In this section the author conducts a sensitivity analysis of the unsprung models by

examining the beam’s internal forces, which are much more sensitive to the number

of modes used when compared with the beam deflection. Recalling Equation (A.16)

and (A.19), the bending moment and shear force of the beam are proportional to the

second and third derivatives of the deflection as follows:

2

2

vM EI

x

∂=

∂ (5.35a)

3

3

vQ EI

x

∂= −

∂ (5.35b)

For the modal solution, the deflection ( ),v x t is computed using Equation (5.10),

which is rewritten again as follows:

( ) ( ) ( ),n n

n

v x t r t xφ=∑ (5.36)

where ( )n xφ is the n-th characteristic shape as defined in Appendix C, and ( )nr t is

the n-th function of time which has to be calculated.


231

With the aid of Equation (C.47) the characteristic shape for a simply supported beam

can be written as follows:

( ) 2 sinn

n xx

L

πφ = (5.37)

The derivatives of Equation (5.37) with respect to x are given as:

( ) ( )( ) 2 cosn x n x L n Lφ π π′ = × (used to find beam rotations) (5.38a)

( ) ( )2

( ) 2 sinn

x n x L n Lφ π π′′ = − × (used to find bending moment) (5.38b)

( ) ( )3

( ) 2 cosn

x n x L n Lφ π π′′′ = − × (used to find shear force) (5.38c)

where each dash over ( )n xφ represents a derivative with respect to x. Comparing

( )n

xφ ′′′ in Equation (5.38c) with 1( )G x′′′ and 2 ( )G x′′′ in Equation (B.8), the reader can see

that the modal derivatives do not suffer discontinuities at the boundaries, unlike the

finite element shape functions. Nevertheless, in Section 3.2.2.4 the author has shown

that the finite element model gives good shear forces using third order derivatives.

Using the same beam and vehicle properties as Section 5.3.3, the author now plots the

vertical displacement, bending moment and shear force at mid-span of the beam as a

function of time using 3, 6 and 12 modes in Figure 5.18. It can be seen from the

results that the beam deflection is insensitive to the number of modes; the bending

moment is slightly sensitive; while the shear force is very sensitive to the number of

modes used.


232

-0.0030

-0.0025

-0.0020

-0.0015

-0.0010

-0.0005

0.0000

0.0005

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

Modal (3 modes)

Modal (6 modes)

Modal (12 modes)

0

50

100

150

200

250

300

350

400

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Be

nd

ing

Mo

me

nt

(kN

m)

Modal (3 modes)

Modal (6 modes)

Modal (12 modes)

-40

-30

-20

-10

0

10

20

30

40

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Sh

ea

r F

orc

e (

kN

)

Modal (3 modes)

Modal (6 modes)

Modal (12 modes)

Figure 5.18: Time history at mid-span of the simply supported beam: (a) vertical

displacement; (b) bending moment; (c) shear force

(a)

(b)

(c)


233

Next, the author compares the bending moment and shear force as functions of time at

mid-span of the beam generated by the modal model and the finite element model as

shown in Figure 5.19. In the finite element model, the beam is discretized into 20

beam elements. Figure 5.19 shows that the bending moments for both systems are

very alike, while the shear force for the modal solution is somewhat better than the

finite element solution with 20 beam elements. Increasing the number of beam

elements in the finite element model will gradually give a better comparison.

0

50

100

150

200

250

300

350

400

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Ben

din

g M

om

en

t (k

Nm

)

Modal (12 Modes)

FEM (20 elements) LN1 at midspan


-40

-30

-20

-10

0

10

20

30

40

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

Time (sec)

Sh

ea

r F

orc

e (

kN

m)

Modal (12 Modes)



Figure 5.19: Comparing the modal with the finite element solution at mid-span of

the beam as a function of time: (a) bending moment; (b) shear force

(a)

(b)


234

Subsequently, the author examines the bending moment and shear force in the simply

supported beam at a specific time i.e. time t = 0.45 sec. Table 5.1 presents the modal

solution for the deflection, bending moment and shear force using Equations (5.37b)

and (5.37c) at a position x = 5 m and 12.5 m at the selected time. From inspection of

the results, it can be seen that the bending moment and shear force at n = 1 differ by

as much as 20% from the bending moment and shear force at n = 12. The deflections

only differ by 2%.

Table 5.1a: Analytical modal bending moment and shear force at x = 5 m

Table 5.1b: Analytical modal bending moment and shear force at x = 12.5 m


235

Figure 5.20 plots the bending moment and shear force at different positions along the

beam at this selected time using the modal model and the finite element method. It

can be seen from the results that the modal solution tends to give slightly lower values

(approximately 5%) than its finite element counterpart. Re-examining Table 5.1, it

would seem unlikely that the bending moment would increase by much beyond the

12th

mode.

0

100

200

300

400

0 5 10 15 20 25

Position along beam (m)

Be

nd

ing

Mo

me

nt

(kN

m)

Modal (12 modes)

FEM (10 elements)

-40

-30

-20

-10

0

10

20

30

40

0 5 10 15 20 25

Position along beam (m)

Sh

ea

r F

orc

e (

kN

)

Modal (12 modes)

FEM (10 elements) LN1

FEM (10 elements) LN2

Figure 5.20: (a) Bending moment; (b) shear force along the simply supported beam

at time t = 0.45 sec for the modal and finite element system

(b)

(a)


236

5.3.6 Multiple unsprung vehicles traversing a simply supported beam

Next, the author examines the effects of several vehicles traversing a simply

supported beam as presented in Figure 5.21. Each vehicle comprises a rigid vehicle

body supported by a pair of axles by means of primary suspensions. The primary

suspension consists of spring and dashpot. The beam has a length L of 30 m, Young’s

modulus of elasticity E of 2.87x107 kN/m


4, mass per

unit length m of 32.4 t/m, Poisson’s ratio ν of 0.2 and first natural frequency 1ω of

25.666 rad/sec (i.e. 4.085 Hz). These properties appear in the works of Yang and Wu

(2001). The bridge beam is discretized into ten beam elements. The data adopted for

the vehicle is the same as that presented by Yang and Wu (2001). This data is taken

from the Manchester vehicle model (Iwnicki, 1999). The unsprung wheel mass Mw is

6.241 t, the vehicle mass Mv is 32 t, the mass moment of inertia of the vehicle Iv is

1970 t m2, the suspension stiffness k1 is 430 kN/m, the suspension damping c1 is 20

kN/m, the distance between axles of a single vehicle Aw is 19 m and the distance

between the rear wheel on a vehicle and the front wheel of the following vehicle Dw is

6 m. The train consists of 10 identical vehicles traversing the beam at a constant speed

c of 27.778 m/s (100km/hr). One ignores both the gravitational and damping effects

of the beam. The Newmark time integration method (Bathe, 1996) with 2000 equal

time steps is used to solve the transient analysis. Time t is arranged in such a manner

that the front wheel is at the left support at t = 0 sec. The initial displacement and

velocity of the unsprung wheel and beam are equal to zero.

The author compares the results from the unsprung finite element system with the

results from the sprung mass system using the ANSYS contact element as well as

with the results using a moving load solution. The Hertzian spring stiffness kH is

1.4x106 kN/m.


237

Figure 5.21: Multiple unsprung vehicles traversing simply supported beam

In Figure 5.22, a plot of the deflection and vertical acceleration at mid-span of the

beam as a function of time for the unsprung finite element system can be seen. It can

be observed from Figure 5.22 that the three models compare well with each other.

-0.0014

-0.0012

-0.0010

-0.0008

-0.0006

-0.0004

-0.0002

0.0000

0.0002

0 2 4 6 8 10

Time (sec)

Vert

ical D

isp

lacem

en

t (m

)

Unsprung f inite element model

Sprung mass system


-0.15

-0.10

-0.05

0.00

0.05

0.10

0.15

0.20

0 2 4 6 8 10

Time (sec)

Ve

rtic

al A

cc

ele

rati

on

(m

/s2)

Unsprung f inite element model

Sprung mass system

Moving Load - Figure 3.30b

Figure 5.22: Time history at mid-span of the beam: (a) vertical displacement; (b)

vertical acceleration

(a)

, v v

M I

wM

1c1k 1k

wM

c

, , E I mwA

L

1c

wD

(b)


238

5.4 Application to the Boyne Viaduct

The author is now confident in analysing the dynamic behaviour of the Boyne

Viaduct railway bridge subjected to the unsprung mass systems. The Boyne Viaduct

is modelled as a simply supported beam during the unsprung modal simulations. The

unsprung finite element system has no restriction regarding the structure shape, thus

the Boyne Viaduct can be represented as a two-dimensional or three-dimensional

structure. More details regarding the modelling of the Boyne Viaduct truss bridge as a

simply supported beam can be found in Appendix F as well as Fryba (2001).

5.4.1 Single unsprung wheel traversing Boyne Viaduct modelled as a beam

The author examines the Boyne Viaduct that is represented as a simply supported

beam, subjected to a single moving unsprung wheel travelling at a slow (10 km/hr)

and fast speed (164 km/hr). In Appendix F, it is found that the Boyne Viaduct can be

represented as a simply supported beam using the following beam properties: length L

of 80.77 m, Young’s modulus of elasticity E of 2.05x108 kN/m

2, moment of inertia I

of 1.22 m4, mass per unit length m of 1.1775 t/m and Poisson’s ratio ν of 0.3. The first

natural frequency 1ω is 22.048 rad/sec, which is equivalent to 3.51 Hz. This natural

frequency differs by about 3% with the natural frequency for the two-dimensional

truss presented in Section 3.2.3. In addition, the unsprung wheel chosen is the front

wheel of a 201 Irish-Rail locomotive. Therefore Mw = 9.3 t, which has an equivalent

weight of 91.25 kN; hence, is 0.1P G . In each example, time t is arranged in such a

manner that the wheel arrives on the bridge at 0ct L = and exits the bridge at

1ct L = . The beam is represented by 12 modes.


239

In Figure 5.23, the author plots the vertical displacement and bending moment at mid-

span of the beam as a function of time. As before, the static effects of the beam are

omitted; hence, only the dynamic effects are taken into account. From inspection of

Figure 5.23, the reader can see that the dynamic response of the beam is minimal at

the slow speed (10km/hr), while at the higher speed (164 km/hr) the beam tends to

experience a noticeable oscillation in deflection and bending moment. In addition, one

can see that modal solution is very similar to the finite element method.

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.2 0.4 0.6 0.8 1.0


De

fle

cti

on

co

eff

icie

nt

Moving Unsprung Wheel - 10 km/hr (Modal)


Moving Unsprung Wheel - 164 km/hr (FEM)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.2 0.4 0.6 0.8 1.0


Be

nd

ing

mo

me

nt

co

eff

icie

nt




Figure 5.23: (a) Vertical displacement; (b) bending moment at mid-span of a simply

supported beam representing the Boyne Viaduct subjected to an

unsprung wheel travelling at 10 km/hr and 164 km/hr

(a)

(b)


240

In Figure 5.24 and 5.25, the author presents the vertical acceleration of the wheel and

contact force between the unsprung wheel and beam as a function of time, where only

the higher vehicle speed is considered. The reaction force between the unsprung

wheel and beam in Figure 5.25 is computed using Equation (5.17a). Inspecting

Figures 5.24 and 5.25, it can be observed that the modal solution and finite element

solution are very similar.

-0.3

-0.2

-0.1

0.0

0.1

0.2

0.3

0.4

0.5

0.0 0.2 0.4 0.6 0.8 1.0


Ve

rtic

al A

cc

ele

rati

on

(m

/s2)



Figure 5.24: Vertical acceleration of the unsprung wheel i.e. riding comfort

-1.10

-1.05

-1.00

-0.95

-0.90

0.0 0.2 0.4 0.6 0.8 1.0


Co

nta

ct

forc

e / S

tati

c w

eig

ht



Figure 5.25: Contact force between the unsprung wheel and beam


241

Before finishing with the modal solution, the author investigates the dynamic effects

of the moving unsprung wheel traversing a simply supported beam (which represents

the Boyne Viaduct) at a wide range of speeds. Since the first natural frequency of the

simply supported beam is 3.51 Hz, the critical speed of the vehicle is computed using

Equation (3.30) as 567 m/s,=cr

c which is 2041 km/hr. As a parametric study, Figure

5.26a plots the dynamic amplification factor DAFU at a range of speeds between

0 1.0α< < for the moving unsprung wheel; however, Figure 5.26b plots the DAFU at

a more realistic range of speeds 10 km/hr to 300 km/hr i.e. 0 0.147α< < for this

structure. A value for α of 0.147 is indicated by the dashed line in Figure 5.26a.

1.0

1.2

1.4

1.6

1.8

2.0

0.0 0.2 0.4 0.6 0.8 1.0


DA

FU

Single Unsprung Wheel (P/G = 0.1)


1.00

1.05

1.10

1.15

1.20

1.25

0 50 100 150 200 250 300

Speed (km/hr)

DA

FU

Single Unsprung Wheel (P/G = 0.1)


Figure 5.26: Dynamic amplification factor at mid-span of the Boyne Viaduct versus

speed ratio: (a) parametric study; (b) realistic vehicle speeds

(a)

(b)


242

Examining Figure 5.26a, it can be seen that the results for the moving unsprung wheel

are similar to the results for the moving load especially at lower speeds 0.25α < ;

however at the higher speed the results can vary by as much as 10%; nonetheless the

curvatures of both sets of results are very similar. Figure 5.26b shows that the

additional wheel inertia causes a slight increase in the dynamic response, which was

also observed in Figure 4.38 for the moving sprung wheel. However, at a range of

speeds between 80 to 100 km/hr, one can see that the results in Figure 5.26b are

different from the results in Figure 4.38. Between these speeds, the moving unsprung

wheel traversing the simply supported beam does not suffer a large increase in the

DAFU, which may only occur when the bridge is modelled as a truss.

5.4.2 Single unsprung wheel traversing two-dimensional Boyne Viaduct

The two-dimensional Boyne Viaduct railway bridge is subjected to an unsprung

wheel travelling at a slow (10 km/hr) and fast speed (164 km/hr). The unsprung wheel

chosen is the front wheel of a 201 Irish-Rail locomotive. Therefore Mw = 9.3 t, which

is equivalent to a weight of 91.25 kN; hence, is 0.1P G . It should be noted that on

this occasion the modal solution is not available as the structure is quite complex to

model. Only solutions from the finite element method developed in Section 5.2.2 are

used. Time t is arranged in such a manner that the wheel arrives on the bridge at

0ct L = and it exits the bridge at 1ct L = .

Figure 5.27(a) and (b) presents the vertical displacement and internal axial forces,

respectively, at mid-span of the bridge as a function of time. The static effects of the

bridge are omitted. Only the dynamic effects are taken into account. From inspection

of Figure 5.27, the reader can see that the dynamic response of the bridge is minimal


243

at the slow speed (10km/hr), while at the higher speed (164 km/hr) the bridge tends to

experience a noticeable oscillation in its deflection and axial force at mid-span. For

comparison purposes, the author also compares the results of the moving unsprung

wheel with the results obtained in Chapter 4 for the moving sprung wheel traversing

the two-dimensional Boyne Viaduct. It can be seen that both the unsprung wheel and

sprung wheel give very similar results. This is due to the reasonably large Hertzian

spring used for the sprung wheel, which has a spring stiffness of 61.4 10 kN/m.×

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.2 0.4 0.6 0.8 1.0


De

fle

cti

on

co

eff

icie

nt

Moving Unsprung Wheel - 10 km/hr


Sprung Wheel - Figure 4.33a - 164 km/hr

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

0.0 0.2 0.4 0.6 0.8 1.0


Ax

ial fo

rce

co

eff

icie

nt



Sprung Wheel - Figure 4.33b - 200 km/hr

Figure 5.27: (a) Vertical displacement; (b) internal axial force at mid-span of the

Boyne Viaduct subjected to an unsprung wheel travelling at 10 km/hr

and 164 km/hr

L

(a)

(b)


244

Next, the vertical acceleration of the unsprung wheel as a function of time is shown in

Figure 5.28. The contact force between the unsprung wheel and bridge as a function

of time is shown in Figure 5.29. Again the author compares these results with the

results from the moving sprung wheel model and it can be seen that the unsprung

wheel tends to be slightly more responsive than the sprung wheel.

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

0.0 0.2 0.4 0.6 0.8 1.0


Ve

rtic

al A

cc

ele

rati

on

(m

/s2)


Sprung Wheel - Figure 4.34 - 164 km/hr

Figure 5.28: Vertical acceleration of the unsprung wheel i.e. riding comfort

-1.2

-1.1

-1.0

-0.9

-0.8

-0.7

0.0 0.2 0.4 0.6 0.8 1.0


Co

nta

ct

forc

e / S

tati

c w

eig

ht


Sprung Wheel - Figure 4.35 - 164 km/hr

Figure 5.29: Contact force between the unsprung wheel and beam


245

Before analysing the unsprung wheel at a range of realistic speeds, the author firstly

conducts a parametric study of the unsprung wheel at a range of speeds, 0 1.0α< < .

In Section 3.2.3 the critical speed of the vehicle cr

c is computed as 550 m/s, which is

equal to 1980 km/hr for the two-dimensional Boyne Viaduct. The dynamic

amplification factors DAFU and DAFA for the unsprung wheel at a range of speed,

0 1.0α< < , can be seen in Figure 5.30. Examining the results, the reader can see that

good similarities can be observed between all results at slow speed, while at higher

speeds the results for the unsprung wheel diverge from the results for the sprung

wheel.

1.0

1.2

1.4

1.6

1.8

2.0

0.0 0.2 0.4 0.6 0.8 1.0

DA

FU


Unsprung Wheel - P/G ratio = 0.1

Sprung Wheel - Figure 4.37a

Moving Load

1.0

1.5

2.0

2.5

3.0

3.5

0.0 0.2 0.4 0.6 0.8 1.0

DA

FA


Unsprung Wheel - P/G ratio = 0.1 (Bot)

Sprung Wheel - Figure 4.37b (Bot)

Moving Load (Bot)

(a)

(b)


246

1.0

1.5

2.0

2.5

3.0

3.5

0.0 0.2 0.4 0.6 0.8 1.0

DA

FA


Unsprung Wheel - P/G ratio = 0.1 (Top)

Sprung Wheel - Figure 4.37b (Top)

Moving Load (Top)



bottom chord; (c) axial force in the top chord

A closer examination of the DAFU at a realistic range of speeds, as represented by

values of α to the left of the vertical dashed line in Figure 5.30a, can be seen in Figure

5.31, which shows that the results for a moving unsprung wheel resemble the results

for the sprung wheel and moving load.

1.00

1.05

1.10

1.15

1.20

0 50 100 150 200 250 300

DA

FU

Speed (km/hr)

Unsprung Wheel - P/G = 0.1

Sprung Wheel - Figure 4.38


Figure 5.31: A close-up view of Figure 5.30a using a realistic range of speeds

(c)


247

5.4.3 Multiple vehicles traversing two-dimensional Boyne Viaduct

This section analyses the vertical response of the two-dimensional Boyne Viaduct

subjected to several moving vehicles. All the wheels are modelled as unsprung

masses. This differs from Section 4.4.3 where the wheels are assumed to be separated

from the rail by a Hertzian spring. Authors such as Chatterjee et al. (1994), Lee

(1998), Yang et al. (1997) and Yang & Wu (2001) have conducted their studies using

the unsprung system, while authors such as Esveld (1989), Zhai & Cai (1997), Zhang

et al. (2001) and Sun & Dhanaekar, 2002 are in favour of the sprung system. In this

section the author compares results from both systems. Vehicles must undergo a static

analysis prior to the execution of the transient analysis. Otherwise sprung components

of the vehicle, i.e. bogie and car body, will not have settled under their own weight at

time t = 0 sec and would go into free vibration as the transient analysis is initiated.

The train model adopted is the same as Section 4.4.3 i.e. a locomotive and three

railway coaches. Figure 5.32(a) and (b) presents the vertical displacement and internal

axial force in the top and bottom chord of the two-dimensional Boyne Viaduct

subjected to the moving train at a slow (10 km/hr) and fast speed (164 km/hr).

Examining Figure 5.32, it can be seen that the dynamic responses of the bridge to the

unsprung wheels and to the sprung wheel models are similar. The vertical

acceleration, i.e. riding comfort, of the locomotive (V1) and first railway coach (V2)

travelling at the faster 164 km/hr is shown in Figure 5.33. The reader should refer to

Figure 4.40 for the position (V1) and (V2) selected on the train. This plot shows that

the absolute value of the vertical acceleration of the locomotive and railway coach is

less than the limit of 1.0 m/s2 recommended in Eurocodes (1990). Comparing Figure

5.33 with Figure 4.42, the reader can see that vertical acceleration of the unsprung

wheel is less than the sprung model; nevertheless, the results are fairly similar.


248

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0


De

fle

cti

on

co

eff

icie

nt

2D Boyne - 10 km/hr (unsprung w heels)


Figure 4.41a - 164 km/hr (sprung w heels)

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

0.0 0.5 1.0 1.5 2.0


Ax

ial fo

rce

co

eff

icie

nt



Figure 4.41b - 164 km/hr (sprung w heels)

Figure 5.32: (a) Vertical displacement; (b) axial force at mid-span of the 2D Boyne


-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

0.0 0.5 1.0 1.5 2.0


Ve

rtic

al A

cc

ele

rati

on

(m

/s2)

2D Train using unsprung w heels - 164 km/hr (V1)


Figure 5.33: Vertical acceleration (riding comfort value) of the 2D vehicle bodies

(b)

Bottom chord

Top chord

(a)


249

In Figure 5.34, the contact forces that exist between the 1st and 7

th unsprung wheels of

the train and the rail are examined at the faster speed (164 km/hr). It can be seen from

the results that as wheels enter the bridge or immediately after they leave the bridge

(and arrive on the rigid rail), the compressive force between the wheel and rail tends

to oscillate abruptly. Figure 5.34 is also somewhat similar to Figure 4.43.

-1.6

-1.4

-1.2

-1.0

-0.8

-0.6

-0.4

0.0 0.5 1.0 1.5 2.0


Co

nta

ct

forc

e / S

tati

c w

eig

ht

2D Train using unsprung w heels - 164 km/hr (W1)




Next, the author examines the dynamic response of the two-dimensional Boyne

Viaduct subjected to the moving train at a range of realistic speeds, 0 0.15α< < i.e.

0 km/hr to 300 km/hr. Again, these results are compared with the results for sprung

wheel train models of Section 4.4.3. Figure 5.35 presents the dynamic amplification

factors DAFU and DAFA for a range of speeds between 10 km/hr to 300 km/hr. From

inspection of Figure 5.35, it can be seen that the amplification factors for unsprung

models tends to be slightly less than those for sprung models at speeds lower than 200

km/hr. The opposite is true at speeds exceeding 200 km/hr.


250

1.00

1.02

1.04

1.06

1.08

1.10

0 50 100 150 200 250 300

Speed (km/hr)

DA

FU

2D Boyne (unsprung w heels)

2D Boyne (sprung w heels) - Figure 4.44a

2D Boyne (moving load)

1.00

1.02

1.04

1.06

1.08

1.10

0 50 100 150 200 250 300

Speed (km/hr)

DA

FA

Bot Chord (unsprung w heels)

Bot Chord (sprung w heels) - Figure 4.44b

Bot Chord (moving load)

1.00

1.02

1.04

1.06

1.08

1.10

0 50 100 150 200 250 300

Speed (km/hr)

DA

FA

Top Chord (unsprung w heels)

Top Chord (sprung w heels) - Figure 4.44c

Top Chord (moving load)



bottom chord; (c) axial force in the top chord.

(a)

(b)

(c)


251

5.4.4 Multiple vehicles traversing three-dimensional Boyne Viaduct

To finish the author examines the dynamic response of the three-dimensional Boyne

Viaduct railway bridge subjected to a train whose wheels are unsprung. The train

travels at operational speeds of 10 km/hr and 164 km/hr. This section also investigates

the dynamic response of the Boyne Viaduct railway bridge at a wide range of vehicle

speeds. Table 5.2 presents the time of execution of the unsprung system as well the

sprung system and multiple moving force system. Each model consists of 1000 equal

time-steps. The program was executed using an Intel Core Duo 2.13 GHz processor

with 2048 RAM. It is clear from the results that the moving force system has the

smallest execution time, while the unsprung system is approximately 25% faster than

the system using ANSYS contact elements and 350% faster than the system using

WRC element. The sprung wheel models require a large number of contact elements,

whether provided by the author or by ANSYS.

Table 5.2: Time of execution for the developed systems

3D Boyne Viaduct with 1000 time steps Cumulative Element Count

Time of execution (sec) Bridge Train Contacts

Train modelled with unsprung wheels

Unsprung wheel-rail contact element system 154 372 - 368

Train modelled with sprung wheels

Wheel-rail contact element system 154 372 788 1372

ANSYS contact elements 154 372 804 464

Train modelled by multiple moving forces

Simple model with overlapping 154 - - 55

Simple model without overlapping 334 - - 40

The vertical displacement, axial force in the top and bottom chord and bending

moment at mid-span of the cross-beam at mid-span of the bridge as a function of time

can be seen in Figure 5.36. It can be seen that there is little difference between the

results for the train modelled with unsprung wheels and those for the train with sprung

wheels.


252

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0


De

fle

cti

on

co

eff

icie

nt



Figure 4.48a - 164 km/hr (sprung w heels)

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

0.0 0.5 1.0 1.5 2.0


Ax

ial fo

rce

co

eff

icie

nt



Figure 4.48b - 164 km/hr (sprung w heels)

0.0

0.2

0.4

0.6

0.8

1.0

1.2

0.0 0.5 1.0 1.5 2.0


Be

nd

ing

mo

me

nt

co

eff

icie

nt 3D Boyne - 10 km/hr (unsprung w heels)


Figure 4.48c - 164 km/hr (sprung w heels)

Figure 5.36: (a) Vertical displacement; (b) axial force; (c) bending moment of the

cross-beam located at mid-span of the 3D Boyne Viaduct due to

vehicles travelling at 10 km/hr and 164 km/hr

Bottom chord

Top chord

(a)

(b)

(c)


253

Next, the vertical acceleration of the locomotive (V1) and first railway coach (V2) as

they travel along the bridge are shown in Figure 5.37. Inspecting Figure 5.37, the

reader can see that the vertical acceleration, in absolute value, or riding comfort of the

locomotive (V1) and railway coach (V2) is less than the 1.0 m/s2 recommended in

Eurocodes (1990). Figure 5.38 presents the contact force between the 1st and 7

th

unsprung wheel and the rail with the train travelling at the faster speed (164 km/hr). It

can be seen from the results that as the wheels enter and exits the bridge; they

experience an abrupt tensile and compressive contact force and oscillate momentarily.

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

0.0 0.5 1.0 1.5 2.0


Ve

rtic

al A

cc

ele

rati

on

(m

/s2)



Figure 5.37: Vertical acceleration (riding comfort) of the 3D vehicle bodies

-3.0

-2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.0 0.5 1.0 1.5 2.0


Co

nta

ct

forc

e / S

tati

c w

eig

ht




th unsprung wheel and the rail


254

Finally, the author investigates the dynamic response of the three-dimensional Boyne

Viaduct subjected to the moving train at a range of realistic speeds, 0 0.15α< < i.e.

0 km/hr to 300 km/hr. These results are compared with the results for sprung wheel

train models of Section 4.4.5 as well as with the results for the two-dimensional

model of Section 5.4.3. In Figure 5.39, the dynamic amplification factors DAFU and

DAFA for a range of speeds between 0 km/hr to 300 km/hr are plotted. Examining

Figure 5.39, it can be seen that the results for the 3D unsprung model are close to the

results for the sprung model. Nevertheless, the results are reasonable similar, with

most results lying within 20% of each other at any particular speed.

1.00

1.02

1.04

1.06

1.08

1.10

1.12

1.14

1.16

0 50 100 150 200 250 300

Speed (km/hr)

DA

FU

3D Boyne (3D Train unsprung w heels)

2D Boyne (2D Train unsprung w heels) - Figure 5.35a

3D Boyne (3D Train sprung w heels) - Figure 4.51a

3D Boyne (3D Train as moving loads) - Figure 3.46

1.00

1.02

1.04

1.06

1.08

1.10

1.12

0 50 100 150 200 250 300

Speed (km/hr)

DA

FA

Bot Chord (3D Train unsprung w heels)

Bot Chord (2D Train unsprung w heels) - Figure 5.35b

Bot Chord (3D Train sprung w heels) - Figure 4.51b

(a)

(b)


255

1.00

1.02

1.04

1.06

1.08

1.10

1.12

0 50 100 150 200 250 300

Speed (km/hr)

DA

FA

Top Chord (3D Train unsprung w heels)

Top Chord (2D Train unsprung w heels) - Figure 5.35c

Top Chord (3D Train sprung w heels) - Figure 4.51b



bottom chord; (c) axial force in the top chord.


To summarize, this chapter develops both a modal model as well as finite element

model for moving unsprung wheels traversing a bridge. The system requires that the

unsprung wheel has the same vertical position as the vertical position of a point of the

beam directly underneath. Since the moving unsprung wheel is moving horizontally,

its vertical velocity is not the same as the vertical velocity of the point of the beam

directly underneath. In fact the vertical velocity of the moving unsprung mass is equal

to the local vertical velocity of the beam plus a convective term. Furthermore, the

vertical acceleration of the moving unsprung mass is equal to the local vertical

acceleration of the beam plus an additional convective term. Wheel-rail separation can

also occur; however, this concept is problematic. As the unsprung wheel regains

contact with the beam, it is likely to cause a large impact load to the beam. A spring

(c)


256

beneath the wheel would resolve this issue, but this idea is already developed in

Chapter 4.

As stated in Section 5.3.2, Akin & Mofid (1989) omit this additional convective

acceleration; thus, their solution for the deflection of the free-end of a cantilever beam

subjected to a moving unsprung mass is inaccurate. Section 5.3.2 also shows that the

results for an unsprung wheel and sprung wheel are comparable provided that the

wheel-rail contact (WRC) element is given a reasonably large Hertzian stiffness and

separation is not allowed. Examining the results presented in Section 5.3.1 to 5.3.5,

the reader can see that the modal model gives very similar results to the finite element

model for a single moving unsprung wheel. In Section 5.3.5, the reader can see that

the deflection at mid-span of the simply supported beam subjected to a moving

unsprung wheel is accurate with only 3 modes. However, at least 12 modes are

required to accurately describe its shear force. Furthermore, the internal forces in the

beam for the modal and finite element models can differ by approximately 5%.

Next, the author analyses the Boyne Viaduct railway-bridge modelled as a simply

supported beam, two-dimensional truss and three-dimensional bridge subjected to

either a single or multiple unsprung wheels. For the Boyne Viaduct modelled as a

simply supported beam (Section 5.4.1), the dynamic amplification factor for the

moving unsprung wheel and for the moving load are similar, especially in the range of

realistic vehicle speeds. When the Boyne Viaduct is modelled as a two-dimensional

truss (Section 5.4.2), the dynamic amplification factor for the moving unsprung

wheel, for a sprung wheel and for a moving load at the range of realistic vehicle

speeds are similar, but quite different at higher speeds beyond 0.2α ≥ .


257

Analyzing the two-dimensional truss (Section 5.4.3) subjected to multiple vehicles;

one found again that the dynamic amplification factor for the unsprung wheel system

and the sprung wheel system are similar. The moving multiple loads described in

Section 3.3.3.2 tend to have a much larger dynamic amplification factor, especially at

high speeds. Examining the results, the DAFU tends to peak at 200 km/hr with an

approximate value of 1.05. Finally in Section 5.4.4, the author simulates the dynamic

response of the three-dimensional Boyne Viaduct railway bridge subjected to a three-

dimensional train modelled with unsprung wheels. The results presented show that the

DAFU also tends to reach a peak at 200 km/hr with an approximate value of 1.07.

Additionally, it is shown that the vertical acceleration or riding comfort of the train is

unlikely to exceed the recommended value of 1.0 m/s2 as it traverses the bridge at its

maximum speed.

As a final note, it can be concluded that the results of the unsprung systems compare

well to the results from the wheel-rail contact (WRC) element. The computational

time of the unsprung system was found to be approximately 25% faster than the

computational time for the use of ANSYS contact elements and about 350% faster

than the computational time for the use of WRC element in the case of the three-

dimensional Boyne Viaduct subjected to multiple vehicles.


258

Chapter 6 – Wheel rail systems with irregularities

259

Chapter 6

Wheel rail systems with irregularities

6.1 Introduction

Until now, the thesis has assumed smooth wheel-rail conditions. However, rails can

have vertical and lateral deviations, which can lead to a pitching and rolling motion of

the vehicle as well as to wheel separation. This chapter goes beyond the scope of the

ANSYS contact elements by modelling rail irregularities for both the sprung and

unsprung wheel.

In the following development the author uses a deterministic irregularity function h

that is defined by the summation of several sinusoidal curves. This is the same

irregularity function as that presented by Chang & Lin (1996) and Yau et al. (1999).

( )1

2 sin

N

j

j j

xh x a

l

π

=

=

∑ (6.1)

where j

l and aj are the wavelength and amplitude, respectively, of the jth

wave, N is

the number of sine curves and x is the horizontal distance from the left support of the

beam.


260

In order to simulate the irregularity function in the wheel-rail contact (WRC) element,

one must apply forces and moments to the nodes of the beam element directly beneath

the wheel, and one must also apply a nodal force to the centre of the wheel. These

forces and moments are related to the beam element shape functions as well as the

Hertzian spring stiffness. The unsprung systems follow a similar development;

however, the vertical acceleration of the unsprung wheel introduces additional terms

such as the second derivative with respect to x for the irregularity function. The author

also develops the wheel-rail contact (WRC) element incorporating lateral rail

irregularities.

To facilitate the validation of the irregularity system, the author makes use of the

Manchester Benchmark Report (Iwnicki, 1999) by comparing and contrasting results

from the developed system with the results from this study. In particular, one

examines case studies TC3 and TC4 from the report, which simulates a freight wagon

(Benchmark Vehicle 2) traversing track with lateral and vertical irregularities.

This then leads to a brief examination of the dynamic effects caused by periodic

irregularities along the Boyne Viaduct. The author examines several scenarios; short

and long wavelength irregularities with in-phase rail irregularities (inducing a vehicle

pitching motion), out-of-phase rail irregularities (inducing a vehicle rolling motion)

and a random irregularity created by several sinusoidal functions. The author finishes

by simulating track with lateral irregularities.


261

6.2 Development of Irregularity models

6.2.1 Sprung wheel incorporating rail irregularities – Finite element model

Similar to Section 4.2.1, the wheel can be modelled as a sprung mass with a Hertzian

spring located between the wheel centre and the rail, as illustrated in Figure 6.1. In

addition, it is assumed that the spring always remains perpendicular to the surface

(Bowe and Mullarkey, 2005).

Figure 6.1: Wheel modelled as a sprung mass traversing an uneven rail

Referring to Figure 4.2a, one can see the sprung mass between nodes i-1 to node i and

its free-body diagram showing irregularities on the rail in Figure 6.2. This figure

shows the Hertzian spring following the profile of the irregularities on the rail. Both

the distance from the left-hand support x and the irregularity function h are shown in

this figure. From inspection of Figure 6.2, one can modify Equation (4.4) as:

( ) ( ) ( ) ( )1 1Ly B B iU v h x v ct x h ctχ −= + = − + (6.2)

Figure 6.2: Free-body diagram of the WRC element with irregularities

Hertzian

Spring

⇒Sprung mass

Wheel Centre

1

1 1

2

2

)(xh

z

x

y

0

Rail

Wheel Centre

2LxU

2LyU

1u

1ˆzθ

1v2

ˆzθ

2u

2v

1

1i −i

x ct=

χ


262

Substituting Equation (6.2) into Equation (4.5) gives the following equation:

( ) ( )2 1

1 12 11 2 12

2 22 21 2 22

2 2

Lx Lx

B i LyL L

Lx LxL L

Ly Ly

U F

v ct x h ct F

U F

U F

−

− +

=

K K

K K (6.3)

Modifying Equation (4.7b) to include the effects of irregularities on the rails yields

the following equation:

( )( )( )( )

( )( )( )( )

[ ]( ) ( )

1 1 1 1

11 1 1 1

1

22 1 2 1

2 1 2 1

i i

B ii i

Ly H H

Lyi i

i i

N ct x N ct x

v ct x h ctG ct x G ct xF k k

UN ct x N ct x

G ct x G ct x

− −

−− −

− −

− −

− −

− + − − = −

− − − −

( )( )( )( )

[ ]( )

( )( )( )( )

[ ]( )

1 1 1 1

11 1 1 1

22 1 2 1

2 1 2 1

0

i i

B ii i

H H H H

Lyi i

i i

N ct x N ct x

v ct xG ct x G ct x h ctk k k k

UN ct x N ct x

G ct x G ct x

− −

−− −

− −

− −

− −

−− − = − + −

− − − −

( )( )( )( )

[ ]( )

( )( )( )( )

( )

1 1 1 1

11 1 1 1

22 1 2 1

2 1 2 1

i i

B ii i

H H H

Lyi i

i i

N ct x N ct x

v ct xG ct x G ct xk k k h ct

UN ct x N ct x

G ct x G ct x

− −

−− −

− −

− −

− −

−− − = − +

− − − −

(6.4)

where the irregularity function ( )h ct has been isolated. Examining the second term on

the right hand side of Equation (6.4), it can be concluded that the effects of the

irregularities are simulated by applying forces and moments to the nodes of a

particular beam segment. The nodal forces and moments are the product of the

Hertzian spring stiffness, the irregularity height and the element shape function. Next

by modifying Equation (4.12c) to include irregularities, one gets:


263

[ ][ ] [ ]

1

1

2 1 1 2 2 2

2

2

0 0

0 0

ˆ0 0

0 0

ˆ1 1

z

Ly H H Ly

z

v

F k N G N G k Uv

θ

θ

= − +

− −

[ ] ( )

0

0

0

0

1

Hk h ct

+ −

−

(6.5)

Again, inspecting the second term on the right-hand side of Equation (6.5), it can be

concluded that the effects of the irregularities also require that a nodal force be

applied to the wheel, which is the product of the Hertzian spring stiffness and the

irregularity height.

The Hertzian extension must also be modified to include the effects of the irregularity

function h and takes the form of Equation (6.6), which is found by substituting

Equation (6.2) into (B.109b), where α is equal to 0o


for the

vertical spring. The contact element of ANSYS is unsuitable for modelling

irregularities as its equation for extension does not include function h.

[ ] [ ]1 2

1 2


Ly Ly

U U


= − +

(6.6a)

[ ]( ) ( )

1

1

2

2

extension 0 1 0 1

Lx

B i

Lx

Ly

U

v ct x h ct

U

U

−

− +

= −

(6.6b)

As explained in Chapter 4, a negative extension, i.e. compression, indicates that

contact exists between the wheel and the rail, while a positive extension, i.e. tension,

means that there is no contact; thus all stiffness matrices, forces and moments related

to that particular wheel are set equal to zero when the extension is positive.


264

6.2.2 Unsprung wheel incorporating rail irregularities – Modal model

Figure 6.3 presents an unsprung mass Mw traversing a section of beam at a constant

speed c. As the unsprung mass travels along the beam, it is assumed that it remains in

direct contact with the beam at all times while following the vertical profile of the

irregularities h.

Figure 6.3: Unsprung wheel traversing a beam with irregularities

Recalling Equation (5.5), the differential equation governing the vibration of a beam

subjected to a moving unsprung wheel mass can be expressed as:

( ) ( ) ( )( )( )

4 2 2

4 2 2

, ,w w

v x t v x t d Y tEI m M M g x X t

x t dtδ

∂ ∂+ = − − −

∂ ∂ (6.7)

Next, one modifies Equation (5.6) to include the irregularity function as follows:

( ) ( )( )

( )( )

,x X t x X t

Y t v x t h x= =

= + (6.8)

Equation (6.8) is the vertical position of the unsprung mass, which is the beam

deflection plus the irregularity function. The vertical velocity of the unsprung mass

with irregularities can then be written as follows:

Mw

z

x

( )dX tc

dt=

y

h


265

( ) ( )( )

( )( )

( ) ( )( )

( ), ,x X t x X t x X t

dY t v x t v x t dX t dh x dX t

dt t x dt dx dt= = =

∂ ∂= + +

∂ ∂ (6.9)

Furthermore, its vertical acceleration is equal to:

( ) ( )( )

2 2

2 2

,x X t

d Y t v x t

dt t=

∂=

∂

( )( )

( ) ( )( )

( ) ( ) ( )( )

( )2 2 2

2 2

, , ,2

x X t x X t x X t

v x t dX t v x t dX t dX t v x t d X t

x t dt x dt dt x dt= = =

∂ ∂ ∂+ + +

∂ ∂ ∂ ∂

( )( )

( ) ( ) ( )( )

( )2 2

2 2x X t x X t

d h x dX t dX t dh x d X t

dx dt dt dx dt= =

+ + (6.10)

Substituting Equation (6.10) into (6.7) yields

( ) ( ) ( )( )

( )( )

( )4 2 2 2

4 2 2

, , , ,2

w x X t x X t

v x t v x t v x t v x t dX tEI m M

x t t x t dt= =

∂ ∂ ∂ ∂+ = − + +

∂ ∂ ∂ ∂ ∂

( )( )

( ) ( ) ( )( )

( )2 2

2 2

, ,x X t x X t

v x t dX t dX t v x t d X tg

x dt dt x dt= =

∂ ∂+ +

∂ ∂

( )( )

( ) ( ) ( )( )

( )( )( )

2 2

2 2x X t x X t

d h x dX t dX t dh x d X tx X t

dx dt dt dx dtδ

= =

+ + −

(6.11)

which extends Equation (5.9) to include irregularity terms. Equation (6.11) is then

solved by the method of superposition using Equation (5.10), which then leads to the

elimination of the fourth-order derivative of ( )n xφ using Equation (C.7 ) and (C.12b),

and finally both sides of the equation are multiplied by ( )i xφ and integrated along the

length of the beam using Equation (C.21) and (C.22) to give the following equation:

( ) ( ) ( ) ( )( )4

1 1 1

N N N

n n in n in w n n

n n n

EI r t a L m r t L M r t X tδ δ φ= = =

+ = −

∑ ∑ ∑&& &&


266

( )( ) ( )( )

( )( ) ( )( )

2

1 1

2N N

n n n n

n n


dt dtφ φ

= =

′ ′′+ +

∑ ∑&

( )( ) ( )( )

2

21

N

n n

n

d X tr t X t g

dtφ

=

′+ +∑

( ) ( )( )

( ) ( )( ) ( )( )

2 2 2

2 2, = 1, 2, 3, ...

ix X t x X t

dX t d h x d X t dh xX t i N

dt dx dt dxφ

= =

+ +

(6.12)

where 1in





( )( ) ( )( ) ( )1

N

in w i n n

n

mL M X t X t r tδ φ φ=

+∑ &&

( )( )( ) ( )( ) ( )

1

2N

w i n n

n

dX tM X t X t r t

dtφ φ

=

′+

∑ &

( )( )( ) ( )( )

2

4

1

N

in n w i n

n

dX tEIL a M X t X t

dtδ φ φ

=

′′+ +

∑

( )( )( ) ( )( ) ( )

2

2w i n n

d X tM X t X t r t

dtφ φ

′+

( ) ( )( )

( ) ( )( ) ( )( )

2 2 2

2 2, = 1, 2, 3, ...

w ix X t x X t

dX t d h x d X t dh xM g X t i N

dt dx dt dxφ

= =

= − + +

(6.13)


1

2

3

1 0 0 ... 0

0 1 0 ... 0

0 0 1 ... 0

...... ... ... ... ...

0 0 0 ... 1 N

r

r

mL r

r

&&

&&

&&

&&

1 1 1 2 1 3 1 1

2 1 2 2 2 3 2 2

3 1 3 2 3 3 3 3

1 2 3

...

...

...

... ... ... ... ... ...

...

N

N

w N

N N N N N N

r

r

M r

r





+

&&

&&

&&

&&


267

( )

1 1 1 2 1 3 1 1

2 1 2 2 2 3 2 2

3 1 3 2 3 3 3 3

1 2 3

...

...

2 ...

... ... ... ... ... ...

...

N

N

w N

N N N N N N

r

rdX t

M rdt

r





′ ′ ′ ′ ′ ′ ′ ′

′ ′ ′ ′ + ′ ′ ′ ′

&

&

&

&

411

422

433

4

0 0 ... 0

0 0 ... 0

0 0 ... 0

...... ... ... ... ...

0 0 0 ... NN

ra

ra

EIL ra

ra

+

( )

1 1 1 2 1 3 1 1

2 2 1 2 2 2 3 2 2

3 1 3 2 3 3 3 3

1 2 3

...

...

...

... ... ... ... ... ...

...

N

N

w N

N N N N N N

r

rdX t

M rdt

r





′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′

′′ ′′ ′′ ′′ + ′′ ′′ ′′ ′′

( )

1 1 1 2 1 3 1 1

2 1 2 2 2 3 2 22

3 1 3 2 3 3 3 32

1 2 3

...

...

...

... ... ... ... ... ...

...

N

N

w N

N N N N N N

r

rd X t

M rdt

r





′ ′ ′ ′ ′ ′ ′ ′

′ ′ ′ ′ + ′ ′ ′ ′

1

2

3

...

w

N

M g

φ

φ

φ

φ

= −

( ) ( )( )

1

2 22

32

...

w x X t

N

dX t d h xM

dt dx

φ

φ

φ

φ

=

−

( ) ( )( )

1

22

32

...

w x X t

N

d X t dh xM

dt dx

φ

φ

φ

φ

=

−

(6.14)

The first and second derivatives of the irregularity function given in Equation (6.1)

are:

( )1

2 2 cos

N

j

j j j

dh x xa

dx l l

π π

=

=

∑ (6.15a)

( )2

2

21

2 2 sin

N

j

j j j

d h x xa

dx l l

π π

=

= −

∑ (6.15b)

In order to compute the new reaction force under the unsprung moving mass, the

author updates Equation (5.17) with Equation (6.10) as follows:


268

( )( )2

2w

d Y tF t M g

dt

= + =

( )( )

( )( )

( )2 2

2

, ,2w x X t x X t

v x t v x t dX tM g

t x t dt= =

∂ ∂+ +

∂ ∂ ∂

( )( )

( ) ( ) ( )( )

( )2 2

2 2

, ,x X t x X t

v x t dX t dX t v x t d X t

x dt dt x dt= =

∂ ∂+ +

∂ ∂

( )( )

( ) ( ) ( )( )

( )2 2

2 2x X t x X t

d h x dX t dX t dh x d X t

dx dt dt dx dt= =

+

(6.16)

6.2.3 Unsprung wheel incorporating rail irregularities – Finite element model

The author now develops a finite element model which includes an unsprung moving

mass on an irregular rail. A free-body diagram of the unsprung mass traversing a

beam segment is shown in Figure 6.4. The diagram indicates that the coordinate

system where x is positive along the beam element, y is positive upward and z is

positive outwards. The deflection in the y-direction and rotation about the z-axis are

defined by R1 and R2, respectively, at local node 1 of the beam, while R3 and R4 are

the deflection in the y-direction and rotation about the z-axis at local node 2 of the

beam.

Figure 6.4: Unsprung wheel traversing a beam segment with irregularities

One begins by substituting Equation (5.18) into (6.11) giving:

Mw

z

x

( )dX tc

dt=

y

h

R1 R3

R2

R4 0 1 2


269

( ) ( ) ( ) ( )4 4

1 1

iv

n n n n

n n

EI R t x m R t x= =

Φ + Φ∑ ∑ &&

( ) ( )( )( )

( ) ( )( )4 4

1 1

2w n n n n

n n

dX tM R t X t R t X t

dt= =

′= − Φ + Φ

∑ ∑&& &

( )( ) ( )( )

( )( ) ( )( )

2 24 4

21 1

n nn n n

n n

dX t d X tR t X t R t X t g

dt dt= =

′′ ′+ Φ + Φ +

∑ ∑

( ) ( )( )

( ) ( )( ) ( )( )

2 2 2

2 2x X t x X t

dX t d h x d X t dh xx X t

dt dx dt dxδ

= =

+ + −

(6.17)

In order to apply the Galerkin’s method of weighted residuals one multiplies both

sides of Equation (6.17) by ( )i

xΦ , i = 1, 2, 3 and 4, and integrates along the element

length as follows:

( ) ( )( ) ( )( ) ( )4

1 0

( )

l

i n w i n n

n


Φ Φ + Φ Φ +

∑ ∫ &&

( )( )( ) ( )( ) ( )

4

1

2 w i n n

n

dX tM X t X t R t

dt=

′Φ Φ +

∑ &

( ) ( )( )

( )( ) ( )( )2

4

1 0

l

iv

i n w i nn

n


dt=

′′Φ Φ + Φ Φ +

∑ ∫

( )( )( ) ( )( ) ( )

2

2w i n n

d X tM X t X t R t

dt

′Φ Φ

( )( )w iM g X t= − Φ

( ) ( )( )

( ) ( )( ) ( )( )

2 2 2

2 2, = 1, 2, 3, and 4

w ix X t x X t

dX t d h x d X t dh xM X t i

dt dx dt dx= =

− + Φ

(6.18)

Substituting Equation (5.25) into (6.18) gives the following equation as:

( ) ( )( ) ( )( ) ( )4

1 0

( )

l

i n w i n n

n


Φ Φ + Φ Φ +

∑ ∫ &&


270

( )( )( ) ( )( ) ( )

4

1

2w i n n

n

dX tM X t X t R t

dt=

′Φ Φ +

∑ &

( ) ( )( )

( )( ) ( )( )2

4

1 0

l

i n w i nn

n


dt=

′′ ′′ ′′Φ Φ + Φ Φ +

∑ ∫

( )( )( ) ( )( ) ( )

2

2w i n n

d X tM X t X t R t

dt

′Φ Φ

( )( ) ( ) ( ) ( ) ( )1 2 1 2ˆ ˆ ˆ ˆ0 0w i i i i iM g X t Q l Q M l M′ ′= − Φ + Φ + Φ + Φ + Φ

( ) ( )( )

( ) ( )( ) ( )( )

2 2 2

2 2, = 1, 2, 3, and 4

w ix X t x X t

dX t d h x d X t dh xM X t i

dt dx dt dx= =

− + Φ

(6.19)

With the aid of Figure B.3, Equation (6.19) can then be reduced to the following

matrix form:

1 1 1 2 1 3 1 4 1 1 1 2 1 3 1 41 1

2 1 2 2 2 3 2 4 2 1 2 2 2 3 2 42 2

3 1 3 2 3 3 3 4 3 1 3 2 3 3 3 430

4 1 4 2 4 3 4 4 4 1 4 2 4 3 4 44

l

w

R R

R Rm dx M

R

R



∫

&& &&

&& &&

&& &

&&

3

4

R

R

&

&&

( )1 1 1 2 1 3 1 4 1 1 1 2 1 3 1 41

2 1 2 2 2 3 2 4 2 1 2 2 2 3 2 42

3 1 3 2 3 3 3 4 33

4 1 4 2 4 3 4 4 4

2 w

R

dX t RM EI

dt R

R



&

&

&

&

1

2

1 3 2 3 3 3 4 30

4 1 4 2 4 3 4 4 4

l

R

Rdx

R

R

′′ ′′ ′′ ′′ ′′ ′′ Φ Φ Φ Φ Φ Φ Φ

′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′Φ Φ Φ Φ Φ Φ Φ Φ

∫

( ) ( )1 1 1 2 1 3 1 4 1 1 1 2 1 3 1 41

2 2

2 1 2 2 2 3 2 4 2 1 2 2 2 3 2 42

2

3 1 3 2 3 3 3 4 3 13

4 1 4 2 4 3 4 4 4

w w

R

RdX t d X tM M

Rdt dt

R



1

2

3 2 3 3 3 4 3

4 1 4 2 4 3 4 4 4

R

R

R

R

′ ′ ′ Φ Φ Φ Φ Φ Φ

′ ′ ′ ′Φ Φ Φ Φ Φ Φ Φ Φ

( ) ( )( )

( ) ( )( )

11 1 12 2 2

2 1 2 2

2 2

3 3 32

4 4 42

ˆ

ˆ

ˆ

ˆ

w w wx X t x X t

Q

M dX t d h x d X t dh xM g M M

dt dx dt dxQ

M

= =

Φ Φ Φ Φ Φ Φ

= − + − − Φ Φ Φ Φ Φ Φ

(6.20)


271

Comparing Equation (6.14) with Equation (6.20), the reader can clearly see that the

two equations are very similar in the representation of the irregularity function. Since

the irregularity function comprises the summation of several sine curves, the first and

second derivatives of this irregularity function are not discontinuous in the finite

element solution, as illustrated by Equation (6.15).

6.2.4 Sprung wheel incorporating lateral rail irregularities – WRC element

In the following section, the author develops the wheel-rail contact (WRC) element

incorporating lateral rail irregularities. Similar to Figure 4.4b, the lateral spring

element is presented in the x-z plane. Hence, the free-body diagram for the lateral

spring element is the same as Figure 6.2 with a change of notation and the direction of

the nodal moments. The free-body diagram with lateral irregularities on the rail is

shown in Figure 6.5. The updated coordinate system now has x positive along the

beam element, y positive inwards and z positive upwards. The deflection in the x, z

plane and rotation about y-axis are denoted as u, w and y

θ , respectively. The lateral

irregularity function f is positive upwards.

Figure 6.5: Free-body diagram of the WRC element with lateral irregularities

Lateral

Spring

1

1 1

2

2

( )f x y

x

z

0

2LxU

2LzU

1u

1ˆ

yθ

1w 2ˆ

yθ

2u

2w

1

ct

1i − i

χ


272

From inspection of Figure 6.5, one can modify Equation (4.31) as follows:

( ) ( ) ( ) ( )1 1Lz B B iU w f x w ct x f ctχ −= + = − + (6.21)

Substituting Equation (6.21) into Equation (4.28) gives the following equation:

( ) ( )2 1

12 11 2 12 1

22 21 2 22 2

2 2

Lx Lx

B iL L Lz

LxL L Lx

Ly Lz

U F

w ct x f ct F

U F

U F

−

− +

=

K K

K K (6.22)

Modifying Equation (4.34b) to include the effects of irregularities on the rails yields

the following equation:

( )( )( )( )

( )( )( )( )

[ ]( ) ( )

1 1 1 1

1 1 1 1 1

1

2 1 2 1 2

2 1 2 1

i i

i i B i

Lz H H

i i Lz

i i

N ct x N ct x

G ct x G ct x w ct x f ctF k k

N ct x N ct x U

G ct x G ct x

− −

− − −

− −

− −

− −

− − − − − + = −

− − − − − −

( )( )( )( )

[ ]( )

( )( )( )( )

[ ]( )

1 1 1 1

11 1 1 1

22 1 2 1

2 1 2 1

0

i i

B ii i

H H H H

Lyi i

i i

N ct x N ct x

w ct xG ct x G ct x f ctk k k k

UN ct x N ct x

G ct x G ct x

− −

−− −

− −

− −

− −

−− − − − = − + −

− − − − − −

( )( )( )( )

[ ]( )

( )( )( )( )

( )

1 1 1 1

1 1 1 11

2 1 2 12

2 1 2 1

i i

i iB i

H H H

i iLz

i i

N ct x N ct x

G ct x G ct xw ct xk k k f ct

N ct x N ct xU

G ct x G ct x

− −

− −−

− −

− −

− −

− − − − − = − +

− − − − − −

(6.23)

where the lateral irregularity function ( )f ct has been isolated. Examining the second

term on the right hand side of Equation (6.23), it can be concluded that the effects of

the irregularities are simulated by applying forces and moments to the nodes of a


273

particular beam segment. The nodal forces and moments are the product of the lateral

spring stiffness, the irregularity height and the element shape function. Next by

modifying Equation (4.39b) to include irregularities, one gets:

[ ][ ] [ ]

1

1

2 1 1 2 2 2

2

2

0 0

0 0

ˆ0 0

0 0

ˆ

1 1

y

Lz H H Lz

y

w

F k N G N G k Uw

θ

θ

= − − − + − −

[ ] ( )

0

0

0

0

1

Hk f ct

+ − −

(6.24)

Again, inspecting the second term on the right-hand side of Equation (6.24), it can be

concluded that the effects of the irregularities also require that a nodal force be

applied to the wheel, which is the product of the lateral spring stiffness and the

irregularity height.

The extension in the lateral spring must also be modified to include the effects of the

irregularity function f and takes the form of Equation (6.25a), which is found by

substituting Equation (6.21) into (B.109b), where α is equal to 270o

and β is equal to

0o

for the lateral spring. The contact element of ANSYS is unsuitable for modelling

irregularities as its equation for extension does not include function f.

[ ] 1

1

extension cos cos cos sinLx

Lz

U

Uβ α β α

= − −

[ ] 2

2

cos cos cos sinLx

Lz

U

Uβ α β α

+ −

(6.25a)

[ ]( ) ( )

1

1

2

2

extension 0 1 0 1

Lx

B i

Lx

Lz

U

w ct x f ct

U

U

−

− +

= −

(6.25b)


274

6.3 Validating systems with irregularities

In order to validate both the sprung and unsprung models including irregularities, the

author compares the results from the models with suitable results obtained from the

literature. The author also conducts a static analysis of the sprung and unsprung

system on an irregular rigid rail and beam.

The beam and vehicle properties adopted in the following examples from Section

6.3.1 to 6.3.2 are similar to those of Section 5.4.1, which modelled to the centre span

of the Boyne Viaduct Railway Bridge, as a simply supported beam. The beam has a

length L of 80.77 m, a Young’s modulus of elasticity E of 2.05x108 kN/m

2, moment

of inertia I of 1.22 m4, mass per unit length m of 1.1775 t/m and Poisson’s ratio ν of

0.3. Using Equation (C.48) the first natural frequency of the simply supported beam

1ω is 22.048 rad/sec, which is equal to 3.51 Hz. The sprung or unsprung wheel mass

Mw is 9.3 t. In Section 6.3.2, only the sprung wheel is positioned on rigid rails at a

distance of 10 m to the left of the left-hand support, and must negotiate the

irregularities on the rigid rail before it enters the beam. The irregularity function

chosen has a wavelength 1 of 5 m,l while the irregularity height is 10 mm; thus the

amplitude 1 is 5 mma or 0.005 m. This irregularity function used can be seen in

Figure 6.6.

The gravitational and damping effects of the beam are again ignored. The Newmark-β

time integration method with 1000 equal time steps is used to solve the transient

analysis. The initial displacement and velocity of the wheel and beam are equal to

zero. Six modes are used to describe the modal solution. In the finite element system

the beam is discretized into 10 beam elements.


275

6.3.1 Static analysis of beam with irregularities

One begins by conducting a static analysis of the three models. In Figure 6.6 the

author presents the vertical displacement at the centre of the wheel at different

positions along the beam, where the left-hand support of the bridge coincides with the

origin. From the result, it can be clearly seen that both the sprung and unsprung

wheels follow the profile of irregularity on the rigid rail. All models experience

positive displacements when the irregularity function is positive and negative

displacements when the irregularity function is negative.

-0.010

-0.005

0.000

0.005

0.010

0.0 1.0 2.0 3.0 4.0 5.0

Position along rigid rail (m)

Ve

rtic

al D

isp

lac

em

en

t (m

)

Irregularity function

WRC Element - Sprung Wheel

FEM & Modal - Unsprung Wheel

Figure 6.6: Static analysis of developed systems with irregularities along a beam

6.3.2 Transient analysis of beam with irregularities

In this example, the author examines the dynamic effects caused by the sprung and

unsprung wheel traversing a beam, whose upper flange is irregular. The two speeds

chosen are a slow speed c of 2.778 m/s (10 km/hr) and a medium speed c = 22.778

m/s (82 km/hr). The vertical displacement and bending moment at mid-span of the

beam as a function of time due to the wheel traversing the beam with irregularities is

shown in Figure 6.7. The vertical displacement and vertical acceleration as well as


276

contact force between the wheel and beam can be seen in Figure 6.8. From inspection

of Figures 6.7 and 6.8, one can see that all models give very similar results as the

wheel traverses this beam with an irregular surface at 10 km/hr. It can also be seen

from each plot that the wheel enters the beam at 0.124ct L = and exits the beam at

1.124ct L = , as indicated by the vertical dashed lines in each plot.

-0.006

-0.005

-0.004

-0.003

-0.002

-0.001

0.000

0.0 0.2 0.4 0.6 0.8 1.0 1.2


Ve

rtic

al D

isp

lac

em

en

t (m

)


Modal (6 modes) - Unsprung Wheel

Finite Element System - Unsprung Wheel

0

500

1000

1500

2000

0.0 0.2 0.4 0.6 0.8 1.0 1.2


Be

nd

ing

Mo

me

nt

(kN

m)




Figure 6.7: (a) Vertical displacement; (b) bending moment at mid-span of the

beam as a function of time due to a wheel traversing at 10 km/hr

across the beam with irregularities

Wheel arriving on LHS

Wheel arriving on RHS

(a)

(b)


277

-0.015

-0.010

-0.005

0.000

0.005

0.0 0.2 0.4 0.6 0.8 1.0 1.2


Ve

rtic

al D

isp

lac

em

en

t (m

)




-0.20

-0.15

-0.10

-0.05

0.00

0.05

0.10

0.0 0.2 0.4 0.6 0.8 1.0 1.2


Ve

rtic

al A

cc

ele

rati

on

(m

/s2)




-1.10

-1.05

-1.00

-0.95

-0.90

0.0 0.2 0.4 0.6 0.8 1.0 1.2


Co

nta

ct

forc

e / S

tati

c w

eig

ht




Figure 6.8 (a) Vertical displacement of wheel; (b) vertical acceleration of wheel;

(c) the contact force between the wheel and beam due to a wheel

traversing at 10 km/hr across the beam with irregularities

(a)

(b)

(c)


278

Next using the same irregularity function, the author examines the dynamic effects on

the bridge and of the wheel at a medium speed c of 22.778 m/s (82 km/hr). The

vertical displacement and bending moment at mid-span of the beam due to this

particular speed and irregularity function are shown in Figure 6.9, while Figure 6.10

presents the vertical displacement and acceleration of the wheel as well as the contact

force between the wheel and beam as a function of time.

-0.010

-0.008

-0.006

-0.004

-0.002

0.000

0.002

0.0 0.2 0.4 0.6 0.8 1.0 1.2


Ve

rtic

al D

isp

lac

em

en

t (m

)




-500

0

500

1000

1500

2000

2500

3000

3500

0.0 0.2 0.4 0.6 0.8 1.0 1.2


Be

nd

ing

Mo

me

nt

(kN

m)




Figure 6.9: (a) Vertical displacement; (b) bending moment at mid-span of the

beam as a function of time due to a wheel traversing at 82 km/hr

across the beam with irregularities

(a)

(b)


279

-0.015

-0.010

-0.005

0.000

0.005

0.0 0.2 0.4 0.6 0.8 1.0 1.2


Ve

rtic

al D

isp

lac

em

en

t (m

)




-10

-8

-6

-4

-2

0

2

4

6

0.0 0.2 0.4 0.6 0.8 1.0 1.2


Ve

rtic

al A

cc

ele

rati

on

(m

/s2)




-2.50

-2.00

-1.50

-1.00

-0.50

0.00

0.0 0.2 0.4 0.6 0.8 1.0 1.2


Co

nta

ct

forc

e / S

tati

c w

eig

ht




Figure 6.10 (a) Vertical displacement of wheel; (b) vertical acceleration of wheel;

(c) the contact force between the wheel and beam due to a wheel

traversing the beam with irregularities, travelling at 82 km/hr

(a)

(b)

(c)


280

As before, the reader can clearly see that Figures 6.9 and 6.10 again show that all

three developed systems give very similar results as the sprung or unsprung wheel

traverses the irregularity function at 82 km/hr. As a final note, the times of execution

of the above example for all three developed systems incorporating irregularities are

presented in Table 6.1. In each simulation, an Intel Core 2.13 GHz processor with

2048 RAM with 1000 equal time-steps is used. The results show that the unsprung

finite element method (FEM) solves twice as fast as the modal unsprung model, and

approximately three times faster than the sprung finite element model.

Table 6.1 Time of execution for the three developed systems with irregularities

Timesteps Time of execution in sec

WRC Modal FEM

1000 178 124 61

6.3.3 Manchester Benchmark simulations using the three models

In the previous two sections, the author has shown that the three models produce

similar results. One is now in a position to compare the results from these systems

with simulations and results from the literature, in particular the author investigates

the dynamic response of a twin-axle freight wagon (Benchmark Vehicle 2) traversing

a track with lateral or vertical irregularities (Track Case 3 or 4) in accordance with the

Manchester Benchmark for Rail Vehicle Simulation, Iwnicki (1999). The natural

frequencies obtained from the vehicle body of the twin-axle freight wagon using the

modal analysis in ANSYS finite element program are compared with some typical

values from the Manchester Benchmark report in Table 6.2. It can be seen from Table

6.2 that the ANSYS results are within a couple percent of the results from the finite


281

element program Vampire, and these results are comparable with other finite element

results.

Table 6.2 Natural frequencies of vehicle body of the twin-axle freight wagon

used in the Manchester Benchmark study (Iwnicki, 1999)

Mode Frequency (Hz)

ANSYS Vampire Other participants

Lateral oscillation 1.069 1.043 0.90 - 1.05

Bouncing 2.075 2.078 2.08 - 2.27

Pitching 2.164 2.298 2.08 - 2.32

Rolling 2.642 2.656 2.58 - 2.90

Yawing 2.729 2.736 2.60 - 2.72

To explain the different vehicle modes, Figure 6.11 presents all six modes that a

railway vehicle can experience as it travels along a rail. Longitudinal oscillation,

bouncing and lateral oscillation are motions parallel to the x, y and z-axis,

respectively. Rolling, yawing and pitching are rotational motions about the x, y, and z-

axis, respectively.

y

x

y

z

Longitudinal oscillation Rolling

x

y

x

z

Yawing Bouncing

(a)

(b) (e)

(d)


282

Figure 6.11: Six modes that the railway vehicle can experience as it travels along

the rail (a) longitudinal oscillation; (b) bouncing; (c) lateral

oscillation; (d) rolling; (e) yawing; (f) pitching (Iwnicki, 2006)

In this first simulation, the vehicle begins on smooth rails and must travel a distance

of 50 m before it undergoes five lateral sinusoidal irregularities as shown in Figure

6.12. Once the vehicle has traversed the five lateral irregularities, the rails become

smooth again. The lateral irregularities on each rail are in-phase with each other. Both

have a wavelength of 11.88 m and a peak-to-peak height of 31.75 mm. In the

Manchester Benchmark for Rail Vehicle Simulation (Iwnicki, 1999) this track is

known as Track Case 3. For this track, the vehicle traverses the rail at a constant

speed of 22.5 m/s (81 km/hr)

Figure 6.12 Plan view of lateral irregularities – Track Case 3 (Iwnicki, 1999)

x

z

Lateral oscillation

x

y

Pitching

(c) (f)


283

The lateral displacement of the leading wheel relative to the track and the lateral

displacement of the vehicle body as a function of time are plotted in Figures 6.13 and

6.14, respectively. The results presented in Figure 6.13a and 6.14a are obtained from

the author’s developed WRC element model, while the results in Figure 6.13b and

6.14b are taken from Iwnicki (1999) studies.

-50

-40

-30

-20

-10

0

10

20

30

40

50

0 50 100 150

Distance Along Track (m)

La

tera

l D

isp

lac

em

en

t (m

m)

Leading Wheel - WRC

Left Track - WRC

Right Track - WRC

Figure 6.13 Lateral displacement of leading wheelset relative to the track as a

function of time: (a) author’s model; (b) Iwinicki (1999). p. 34, Fig 24

(b)

(a)


284

-6

-4

-2

0

2

4

6

0 50 100 150


La

tera

l D

isp

lac

em

en

t (m

m)

Vehicle Body - WRC

Figure 6.14 Lateral displacement of the vehicle body as a function of time: (a)

author’s models; (b) Iwinicki (1999). p. 34, Fig 23

Examining the results in Figure 6.14, the reader can see that the lateral displacement

of the vehicle body is somewhat different than the results from the literature.

Nevertheless, the lateral displacement of the leading wheelset shown in Figure 6.13

has good agreement with Iwinicki (1999) results. The periodic lateral movement of

axles known as the Klingel movement (Esveld, 1989) has been omitted from the

author’s model, which could explain the differences between Figure 6.14a and 6.14b.

(a)

(b)


285

In this next simulation, the vehicle again begins on smooth rail and must travel a

distance of 50 m before it undergoes any vertical irregularities. Beyond 50 m, from

the starting position, the vehicle is then subjected to vertical sinusoidal irregularities,

with a wavelength of 9 m and peak-to-peak height of 20 mm as shown in Figure 6.15.

During this particular simulation, the vehicle begins with an initial speed of 20 m/s

(72 km/hr) as it travels along the smooth rail for 50 m. Then between the distances 50

to 250 m, from its original starting position, the speed of the vehicle is increased

linearly from 20 to 24 m/s (86.4 km/hr) with an acceleration of 0.44 m/s2. Beyond 250

m, the vehicle travels with a constant speed of 24 m/s for another 120 m. It should be

noted that the acceleration term ( )2 2d X t dt in the unsprung model is activated in

this solution.

Figure 6.15 Elevation view of vertical irregularities – Track Case 4 (Iwnicki, 1999)

The vertical displacement of the vehicle body and the vertical displacement of the

leading wheelset relative to the track as a function of time are plotted in Figures 6.16

and 6.17, respectively. As before, the results in Figure 6.16a and 6.17a are obtained

from the author’s developed models, while the results in Figure 6.16b and 6.17b are

found in Iwnicki (1999) studies. It can be seen from Figures 6.16a and 6.17a that the

results of the moving sprung model with irregularities are very similar to the results of


286

the moving unsprung model with irregularities. In addition, by comparing Figure

6.16a with 6.16b, the reader can see that the results of the author’s models are

comparable with the results obtained from the Manchester Benchmark report.

Moreover, the results in Figures 6.17a and 6.17b behave in a similar manner.

0

3

6

9

12

15

0 50 100 150 200 250 300


Ve

rtic

al D

isp

lac

me

nt

(mm

)

Unsprung Model - FEM

Sprung Model - WRC

Figure 6.16: Vertical displacement at the centre of the vehicle body as a function of

time: (a) author’s models; (b) Iwinicki (1999). p. 37, Fig 29

(a)

(b)


287

0

5

10

15

20

25

0 50 100 150 200 250 300


Ve

rtic

al D

isp

lac

me

nt

(mm

)

Unspung Model - FEM Spung Model - WRC

Figure 6.17: Vertical displacement of the leading wheelset relative to the track as a

function of time: (a) author’s models; (b) Iwinicki (1999). p 37, Fig 30

(a)

(b)


288

6.4 Boyne Viaduct with irregularities along the rail

After establishing that the author’s models are suitable for simulating track

irregularities, the author is now in a position to investigate the effects of a three-

dimensional train traversing the Boyne Viaduct with rail irregularities. To ensure the

bridge does not suffer any initial shock due to the irregularities, the front wheel of the

train is positioned on rigid rail at a distance of 10 m to the left of the left support. This

also allows the train to undergo pitching or rolling motion prior to entering the bridge.

In the following examples, the train comprises a single six-axle 201 Class locomotive

and a single four-axle Mark 4 railway coach. This train model has a computational

time half that of the train model used in Section 4.4.5. In the following sub-sections,

the author analyses the dynamic response of the Boyne Viaduct and railway vehicle

subjected to four types of irregularity function:

1. Single irregularity function, in-phase on each rail, inducing pitching motion

2. Single irregularity function, out-of-phase on each rail, inducing rolling motion

3. Multiple random irregularity functions on each rail

4. Single lateral irregularity function, in-phase on each rail

In addition, the author examines the effects of the train traversing irregularity

functions with short and long wavelengths. In one example, a wavelength 1 of 5l m

(short) is used, while in the other a wavelength 1 of 40l m (long). The height is 10

mm; thus, 1 5 mma = or 0.005 m as shown in Figure 6.18. In each simulation, the

train travels along the uneven track at a constant speed c of 22.78 m/s (82 km/hr),

which is half its maximum operating speed. For convenience, the train is now

modelled using sprung wheels; nevertheless, Section 6.3 has shown that the unsprung

wheel system is an equally suitable method for modelling irregularities.


289

-10

-5

0

5

10

-40 -20 0 20 40 60 80 100 120


Ve

rtic

al P

rofi

le (

mm

)

Short Wave Irregularity - 5 m

Long Wave Irregularity - 40 m

Figure 6.18: Vertical profile of two types of irregularity function considered

6.4.1 Boyne Viaduct with vertical irregularities on each rail in-phase

In this first sub-section, each rail along the Boyne Viaduct is given in-phase

irregularities, thus, inducing a vehicle pitching motion (rotation about the z-axis) as

the vehicle travels along the rail. The two irregularity functions examined are shown

in Figure 6.18, one has a wavelength 1 of 5l m (short) and the other has a wavelength

2 of 40l m (long). The vertical displacement, axial force and bending moment at mid-

span of the Boyne Viaduct as a function of time due to the train traversing both types

of irregularities are shown in Figure 6.19.

-0.030

-0.025

-0.020

-0.015

-0.010

-0.005

0.000

0.005

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75

Dimesionless Tme ct /L

Ve

rtic

al D

isp

lac

em

en

t (m

)

Short Wave Irregularity

Long Wave Irregularity

Left support Right support

Front wheel on LHS

Front wheel on RHS

Rear wheel on RHS

(a)


290

-1500

-1000

-500

0

500

1000

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75

Dimesionless Time ct /L

Ax

ial F

oc

e (

kN

)



0

50

100

150

200

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Be

nd

ing

Mo

me

nt

(kN

m)



Figure 6.19: (a) Vertical displacement; (b) axial force; (c) bending moment (mid-

span of the cross-beam) at mid-span of the Boyne Viaduct as a function

of time due to the train travelling at 82 km/hr across in-phase

irregularities on each rail

In Figure 6.20, the author now plots the vertical displacement of the front wheel of the

train (W1) as well as its wheel-rail contact force, while Figures 6.21 and 6.22 present

the vertical acceleration and pitching angle (rotation about z-axis), respectively, of the

vehicle body of the locomotive (V1) and railway coach (V2) as a function of time as

the train travels along the in-phase irregularities.

(b)

(c)

Bottom Chord

Top Chord


291

-0.04

-0.03

-0.02

-0.01

0.00

0.01

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Ve

rtic

al D

isp

lac

me

nt

(m)



-2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Co

nta

ct

Fo

rce

/ S

tati

c W

eig

ht



Figure 6.20: (a) Vertical displacement of wheel; (b) contact force between the front

wheel of the train and rail as a function of time due to the train

travelling at 82 km/hr over in-phase irregularities on each rail

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Ve

rtic

al A

cc

ele

rati

on

(m

/s2) Short Wave Irregularity - Loco

Long Wave Irregularity - Loco

(a)

(b)

(a)


292

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Ve

rtic

al A

cc

ele

rati

on

(m

/s2) Short Wave Irregularity - Coach

Long Wave Irregularity - Coach

Figure 6.21: Vertical acceleration of the vehicle body as a function of time

subjected to in-phase irregularities: (a) locomotive; (b) railway coach

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Pit

ch

ing

An

gle

(m

rad

)

Short Wave Irregularity - Loco


-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Pit

ch

ing

An

gle

(m

rad

)

Short Wave Irregularity - Coach


Figure 6.22: Pitching angle of the vehicle body as a function of time subjected to in-

phase irregularities: (a) locomotive; (b) railway coach

(a)

(b)

(b)


293

In Figure 6.19, the dynamic response of the bridge structure for the short and long

wave irregularity functions are very similar. The results for the train for the short and

long wave irregularity functions are quite different. Yau et al. (1999) and Zhang et al.

(2001) have noted that the dynamic response of the bridge is not sensitive to

irregularities. The dynamic response of the vehicle is very sensitive to differences in

wavelength and amplitude between irregularity functions.

Studying Figure 6.20a, one can see that the front wheel of the train follows both

irregularity profiles given in Figure 6.18. Figure 6.20b then shows that the shorter

irregularity function tends to produce larger fluctuations in the contact force than the

longer irregularity function. This indicates that for the shorter irregularity the wheel is

more susceptible to separation from the rail at higher speeds. From inspection of

Figure 6.21, it can be seen that the vertical acceleration in absolute value or riding

comfort of the locomotive and railway coach is less than 1.0 m/s2, most of the time, as

recommended in Eurocodes (1990). The pitching angle of the vehicle body, measured

in radians/1000 i.e. (mrad), is shown in Figure 6.22.

Exaggerating the vertical, horizontal and lateral displacements as well as the vertical

height of the irregularity function by a factor 20, the author draws the train traversing

the Boyne Viaduct in the presence of the shortwave irregularity function, at two

different times in Figure 6.23. As shown in the diagrams, the bogies of the vehicles

are more susceptible to pitching than the vehicle body.


294

Figure 6.23: Animation frame taken at (a) 2.425 sec; (b) 3.785 sec for the

locomotive and railway carriage traversing the Boyne Viaduct railway

bridge at 82 km/hr in the presence of short in-phase irregularities (all

displacements and irregularity function exaggerated by a factor of 20)

(a)

(b)

Direction of travel


295

6.4.2 Boyne Viaduct with vertical irregularities on the rail out-of-phase by 180o

The author is now interested in the rolling effects (rotation about the x-axis) of the

vehicle. Thus, the rails have a single sinusoidal irregularity out of phase by 180o with

each other. As before, the two irregularity functions chosen are shown in Figure 6.18.

One has a wavelength 1 of 5l m (short) and the other has a wavelength 2 of 40l m

(long). The right rail is on the right hand side of the observer who is facing the

direction of travel of the train. The left rail has the same irregularity function as

shown in Figure 6.18, while the right rail has the same irregularity function that is

out-of-phase by 180o with the left rail. The vertical displacement, axial force and

bending moment at mid-span of the Boyne Viaduct as a function of time due to the

train traversing both types of irregularities are shown in Figure 6.24.

-0.030

-0.025

-0.020

-0.015

-0.010

-0.005

0.000

0.005

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Ve

rtic

al D

isp

lac

em

en

t (m

)



-1500

-1000

-500

0

500

1000

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Ax

ial F

oc

e (

kN

)



Front wheel on LHS

Front wheel on RHS

Rear wheel on RHS

Bottom Chord

Top Chord

(a)

(b)


296

0

50

100

150

200

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Be

nd

ing

Mo

me

nt

(kN

m)



Figure 6.24: (a) Vertical displacement; (b) axial force; (c) bending moment at mid-

span of the Boyne Viaduct as a function of time due to the train

travelling at 82 km/hr across out-of-phase irregularities on each rail

Next, the author plots the vertical displacement of the front wheel of the train (W1) as

well as its wheel-rail contact force, where Figure 6.25 presents the left front wheel

and Figure 6.26 presents the right front wheel. The vertical acceleration and rolling

angle (rotation about x-axis) of the vehicle body of the locomotive (V1) and railway

coach (V2) as a function of time as the train travels along the out-of-phase

irregularities are then shown in Figures 6.27 and 6.28, respectively.

-0.04

-0.03

-0.02

-0.01

0.00

0.01

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Ve

rtic

al D

isp

lac

me

nt

(m)



(c)

(a)


297

-2.0

-1.5

-1.0

-0.5

0.0

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Co

nta

ct

Fo

rce

/ S

tati

c W

eig

ht



Figure 6.25: (a) Vertical displacement of left wheel; (b) contact force between the

front left wheel and rail of the train as a function of time (left rail)

-0.04

-0.03

-0.02

-0.01

0.00

0.01

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Ve

rtic

al D

isp

lac

me

nt

(m)



-2.0

-1.5

-1.0

-0.5

0.0

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Co

nta

ct

Fo

rce

/ S

tati

c W

eig

ht



Figure 6.26: (a) Vertical displacement of right wheel; (b) contact force between the

front right wheel and rail of the train as a function of time (right rail)

(b)

(a)

(b)


298

-1.0

-0.5

0.0

0.5

1.0

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Ve

rtic

al A

cc

ele

rati

on

(m

/s2) Short Wave Irregularity - Loco


-1.0

-0.5

0.0

0.5

1.0

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Ve

rtic

al A

cc

ele

rati

on

(m

/s2) Short Wave Irregularity - Coach


Figure 6.27: Vertical acceleration of the vehicle body subjected to out-of-phase

irregularities as a function of time: (a) locomotive; (b) railway coach

-15

-10

-5

0

5

10

15

20

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Ro

llin

g A

ng

le (

mra

d)

Short Wave Irregularity (multiplied by a factor of 10)


(a)

(b)

(a)


299

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

2.5

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Ro

llin

g A

ng

le (

mra

d)

Short Wave Irregularity (multiplied by a factor of 10)


Figure 6.28: Rolling angle of the vehicle body subjected to out-of-phase

irregularities as a function of time: (a) locomotive; (b) railway coach

Examining the results in Figure 6.24, one can see that as the train traverses the bridge

over out-of-phase irregularities, the bridge shows little additional dynamic response.

Comparing the results in Figure 6.20 with the results in Figure 6.25 and 6.26, it can be

seen that the wheels tend to correctly follow the vertical profile shown in Figure 6.18.

In addition, the shorter irregularity function gives rise to less response in Figure 6.25b

and 6.26b when compared to Figure 6.20. The vertical acceleration in absolute value

or riding comfort of both vehicle types is then shown to be less than 0.5 m/s2, most of

the time in Figure 6.27. Finally, Figure 6.28 shows that the roll angle of the vehicle

body (V1 and V2) are much larger for the long wave than for the shorter wave

irregularity function.

Exaggerating the vertical, longitudinal and lateral displacements as well as the vertical

height of the irregularity function by a factor 20, the author draws the train traversing

the Boyne Viaduct in the presence of a long wave irregularity function, in Figure 6.29.

As observed in Figure 6.28, the vehicle bodies experience rolling effects.

(b)


300



bridge at 82 km/hr in the presence of short in-phase irregularities (all

displacements and irregularity function exaggerated by a factor of 20)

(b)

(a) Right rail

Left rail

Direction of travel


301

6.4.3 Boyne Viaduct with random vertical irregularities on each rail

Next, the author examines the dynamic responses of both the bridge and train in the

presence of random irregularities on both tracks. The irregularity function used on

each track comprises several sine curves (N = 5); thus, simulating more realistic rail

conditions and can be seen in Figure 6.30 or equated using Equation (6.1) as follows:

Left

2 2 2 2 2 0.002 sin sin sin sin sin

8 16 24 32 40

x x x x xh

π π π π π = − − − +

(6.26a)

Right

2 2 2 2 2 0.002 sin sin sin sin sin

7 13 19 23 37

x x x x xh

π π π π π = + + + +

(6.26b)

It should be again noted that the rail on the observer’s right-hand side as he faces the

train’s direction of travel is denoted the right rail. The other rail is the left rail. As in

previous sections, the train traverses the uneven track at a constant speed c = 22.78

m/s (82 km/hr), which is half its maximum operating speed.

-10

-5

0

5

10

-40 -20 0 20 40 60 80 100 120


Ve

rtic

al P

rofi

le (

mm

)

Left Rail

Right Rail

Figure 6.30: Vertical profile of the random irregularity functions on each rail

The vertical displacement, axial force in the upper and lower chord and bending

moment in the centre of the cross-beam at mid-span of the Boyne Viaduct as a

function of time is shown in Figure 6.31. For comparison purposes, the author plots

on Figure 6.31 the dynamic response of the bridge as the train traverses the bridge in

the presence of a smooth rail.



302

-0.030

-0.025

-0.020

-0.015

-0.010

-0.005

0.000

0.005

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Ve

rtic

al D

isp

lac

em

en

t (m

)

Smooth Rail

Random Irregularities

-1500

-1000

-500

0

500

1000

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Ax

ial F

orc

e (

kN

) Smooth Rail


0

50

100

150

200

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Be

nd

ing

Mo

me

nt

(kN

m)

Smooth Rail



span of the Boyne Viaduct as a function of time. The train is travelling

at 82 km/hr across random irregularities on each rail

Front wheel on LHS

Front wheel on RHS

Rear wheel on RHS

Bottom Chord

Top Chord

(a)

(b)

(c)


303

Next, the author plots the vertical displacement of the front wheels of the train (W1)

as well as the wheel-rail contact forces as a function of time in Figure 6.32. The

vertical acceleration, pitching angle and rolling angle of the vehicle body of the

locomotive (V1) and railway coach (V2) as a function of time are then shown in

Figures 6.33a, 6.33b and 6.33c, respectively.

-0.05

-0.04

-0.03

-0.02

-0.01

0.00

0.01

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Ve

rtic

al D

isp

lac

me

nt

(m)

Wheel on Left Rail

Wheel on Right Rail

-2.0

-1.5

-1.0

-0.5

0.0

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Co

nta

ct

Fo

rce

/ S

tati

c W

eig

ht

Wheel on Left Rail

Wheel on Right Rail

Figure 6.32: (a) Vertical displacement of front wheels; (b) contact force between the

fronts wheel and rail of the train as a function of time

(a)

(b)


304

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Ve

rtic

al A

cc

ele

rati

on

(m

/s2)

Loco Coach

-3.0

-2.0

-1.0

0.0

1.0

2.0

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Pit

ch

ing

An

gle

(m

rad

)

Loco

Coach

-3.0

-2.0

-1.0

0.0

1.0

2.0

3.0

0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75


Ro

llin

g A

ng

le (

mra

d)

Loco Coach

Figure 6.33: (a) Vertical acceleration; (b) pitching angle; (c) rolling angle of the

vehicle body as a function of time. The train is travelling at 82 km/hr

across random irregularities on each rail

(a)

(b)

(c)


305

From inspection of Figure 6.31, the reader can see that the observation of Yau et al.

(1999) and Zhang et al. (2001) are true, namely that the dynamic response of the

bridge is not affected by irregularities along its rails. Figure 6.32 then shows the front

wheels of the train following the vertical profile of each rail presented in Figure 6.30,

along with the wheel-rail contact forces. Unlike previous examples, the vertical

acceleration, in absolute value, or riding comfort value of the locomotive and railway

coach, tends to barely lie within the recommended value of 1.0 m/s2 on most

occasions in Figure 6.33a. Figure 6.33b and 6.33c show that the random irregularities

induce pitching and rolling of the vehicle bodies. The locomotive tends to experience

a greater roll angle than the railway coach. Finally, Figure 6.34 presents two drawings

at different times of the train traversing the Boyne Viaduct over random irregularities

along each rail. All bridge and vehicle displacements as well as the irregularity

function have been exaggerated by a factor of 20.

(a)

Right rail

Left rail

Direction of travel


306



bridge at 82 km/hr undergoing random rail irregularities on each rail

(displacements and irregularity function exaggerated by a factor of 20)

6.4.4 Boyne Viaduct with lateral irregularities in-phase on each rail

In this final sub-section, the author examines the dynamic response of a train

traversing a track with lateral irregularities in-phase on each rail; thus, inducing a

lateral oscillation of the vehicle body. The lateral irregularity function used has a

wavelength of 1 of 40l m and a peak-to-peak height of 50 mm; hence its amplitude 1a

is 25 mm or 0.025m as illustrated in Figure 6.35. As in previous sections, the train

traverses the uneven track at a constant speed c = 22.78 m/s (82 km/hr), which is half

its maximum operating speed.

(b)


307

-50

-25

0

25

50

-40 -20 0 20 40 60 80 100 120


La

tera

l P

rofi

le (

mm

)

Figure 6.35: Lateral profile of irregularity function considered (Plan view)

For consistency, the vertical displacement, axial force in the upper and lower chord

and bending moment in the centre of the cross-beam at mid-span of the Boyne

Viaduct as a function of time is shown in Figure 6.36. For comparison purposes again,

the dynamic response of the bridge without lateral irregularities is also shown in

Figure 6.36. Comparing Figure 6.36 with Figures 6.19, 6.24 and 6.31, the reader can

see that the lateral irregularities on the track have little dynamic effect on the Boyne

Viaduct, similar to previous observations.

-0.030

-0.025

-0.020

-0.015

-0.010

-0.005

0.000

0.005

0 0.25 0.5 0.75 1 1.25 1.5 1.75

Dimensionless Time ct /L

Ve

rtic

al D

isp

lac

em

en

t (m

)

Smooth Rail

Lateral Irregularity


(a)


308

-1500

-1000

-500

0

500

1000

1500

0 0.25 0.5 0.75 1 1.25 1.5 1.75


Ax

ial F

orc

e (

kN

)Smooth Rail


0

50

100

150

200

0 0.25 0.5 0.75 1 1.25 1.5 1.75


Be

nd

ing

Mo

me

nt

(kN

m)

Smooth Rail



span of the Boyne Viaduct as a function of time. The train is travelling

at 82 km/hr along the track with in-phase lateral irregularities

Next, the author plots the lateral displacement of the front wheels of the train (W1) as

well as the wheel-rail contact forces as a function of time in Figure 6.37. The vertical

acceleration, lateral acceleration, rolling angle (see Figure 6.11) and yawing angle of

the vehicle body of the locomotive (V1) and railway coach (V2) as a function of time

are then shown in Figures 6.38a, 6.38b, 6.38c and 6.33d, respectively.

Bottom Chord

Top Chord

(c)

(b)


309

-0.050

-0.025

0.000

0.025

0.050

0 0.25 0.5 0.75 1 1.25 1.5 1.75


La

tera

l D

isp

lac

em

en

t (m

)

Front w heel of the locomotive (W1)

Front w heel of the railw ay coach (W7)

-2.0

-1.5

-1.0

-0.5

0.0

0 0.25 0.5 0.75 1 1.25 1.5 1.75


Co

nta

ct

Fo

rce

/ S

tati

c W

eig

ht

Front w heel of the locomotive (W1)

Front w heel of the railw ay coach (W7)

Figure 6.37: (a) Lateral displacement; (b) vertical contact force between the 1st and

7th

wheel and the rail

-1.00

-0.50

0.00

0.50

1.00

0 0.25 0.5 0.75 1 1.25 1.5 1.75


Ve

rtic

al A

cc

ele

rati

on

(m

/s2)

Loco (V1)

Coach (V2)

(a)

(b)

(a)


310

-1.00

-0.50

0.00

0.50

1.00

0 0.25 0.5 0.75 1 1.25 1.5 1.75


La

tera

l A

cc

ele

rati

on

(m

/s2)

Loco (V1)

Coach (V2)

-6.0

-4.0

-2.0

0.0

2.0

4.0

6.0

0 0.25 0.5 0.75 1 1.25 1.5 1.75


Ro

llin

g A

ng

le (

mra

d)

Loco (V1)

Coach (V2)

-6.0

-4.0

-2.0

0.0

2.0

4.0

6.0

0 0.25 0.5 0.75 1 1.25 1.5 1.75


Ya

win

g A

ng

le (

mra

d)

Loco (V1)

Coach (V2)

Figure 6.38: (a) Vertical acceleration; (b) lateral acceleration; (c) rolling angle;

(d) yawing angle of the vehicle body as a function of time. The train is

travelling at 82 km/hr across lateral irregularities on each rail

(d)

(c)

(b)


311

From inspection of Figure 6.37, one can see that the wheels of the train (W1 and W7)

correctly follow the lateral profile as presented in Figure 6.35. In addition, these

wheels remain in contact with the rails at all times, oscillating somewhat close to its

own static weight. The vertical and lateral acceleration, in absolute value, or riding

comfort value of the locomotive and railway coach, tends to lie well within the

recommended value of 1.0 m/s2 on all occasions as shown in Figure 6.38a and Figure

6.38b. Figures 6.38c and 6.38d show that the lateral irregularities induce rolling and

yawing of the vehicle bodies.


The main purpose of this chapter is the modification of the sprung and unsprung

wheel-rail systems to incorporate track irregularities. ANSYS contact elements cannot

incorporate such irregularities. The author uses a deterministic irregularity function

defined by one or more sinusoidal curves to simulate uneven rail conditions. For the

sprung finite element system, it is shown that the irregularity function along the track

can be simulated for each wheel by applying a force to the wheel and forces and

moments to the nodes of the element underneath the wheel. In the unsprung system,

the acceleration terms in Equation (6.14) to (6.20) involve the second derivative of the

irregularity function.

Examining the results in Section 6.3.1 to 6.3.3, the reader can see that the sprung and

unsprung wheel systems with irregularities along the track are similar. Furthermore,

the author’s results compare reasonably well with the Manchester Benchmark

simulation i.e. a twin-axle freight wagon traversing a track with lateral or vertical

irregularities (Track Case 3 or 4). Of the three developed systems with irregularities,


312

the unsprung finite element model has the fastest computational time. Its

computational time is half of the modal solution and a third of the wheel-rail contact

(WRC) element. Despite this slow time, it is the preferred method used by the author

to simulate a train traversing the Boyne Viaduct with track irregularities, as it is able

to simulate wheel-rail separation.

The results presented in Section 6.4, for the Boyne Viaduct, are in good agreement

with the literature of Yau et al. (1999) and Zhang et al. (2001) that the dynamic

response of the bridge due to the presence of track irregularities are insignificant,

while the dynamic response of the vehicle is substantial. This is evident in Figures

6.19, 6.24, 6.30 and 6.36 as the dynamic responses of the Boyne Viaduct show little

change in all the different simulations, while Figures 6.20 to 6.22, 6.25 to 6.28, 6.32

to 6.33 and 6.37 to 6.38 show a significant difference with one another.

As a final note, the drawings presented by the author in Figures 6.23, 6.29 and 6.34,

clearly show the train traversing the different track profiles, which is a primary aim of

this thesis to create wheel-rail contact elements that can simulate track irregularities.

Chapter 7 – Conclusions & Recommendations

313

Chapter 7

Conclusions & Recommendations

7.1 Thesis Summary & Conclusions

This thesis primarily focuses on the dynamic interaction between trains and bridges.

The trains are simulated as moving point loads in Chapter 3. Wheels are presented as

sprung masses in Chapter 4, and as unsprung masses in Chapter 5. In Chapter 6

smooth rails are replaced with rails with irregular upper surfaces. All simulations have

been conducted using the ANSYS finite element program. The ANSYS program has

some limitations. Many of these limitations have been overcome with mathematical

algorithms developed by the author.

Train’s modelled as moving point loads are acceptable if the mass of the bridge is

substantially greater than the mass of the vehicle. This is true of large spanning

bridges (Fryba, 1996). In the moving point load system, nodal forces (simple model)

or nodal forces and nodal moments (exact model) are applied to nodes along the beam

as a function of time to represent the passage of a train. It was found that the results

from the simple model are comparable to the results from the exact model provided

that a sufficient number of beam elements is used to discretize the beam. One did find


314

that the results for the simple model with fewer beam elements were remarkably

similar to the results from the ANSYS contact element, both losing accuracy as the

number of beam elements was reduced. The exact model did not suffer a loss of

accuracy of deflection with a reduction of beam elements. However, it has a

significantly different bending moment. Applying joint equilibrium on a particular

node, one found that the summation of the two internal moments at that node was

equal to the applied moment PG1 or PG2, depending on which element the moving

force was located on. This then gave a stepped bending moment across the beam

element. It was also revealed that that the internal forces from Equation (3.6) and (3.5)

for the simple and exact models are exactly the same internal forces as from a pressure

applied between two consecutive nodes.

The next development involves the wheel-rail contact (WRC) element. It can be

described as an improvement on the moving load system. Springs are placed between

the wheel and rail. These springs can be vertical (Hertzian), longitudinal, or lateral.

The longitudinal spring allows the train to brake or accelerate. The lateral springs

provide lateral stability of the wheels in three-dimensional models. The longitudinal

and lateral features are currently unavailable within ANSYS. Through the results

presented, the reader can see that the WRC element gives remarkably similar results

to the ANSYS contact elements. Additionally, the author has shown that the ANSYS

contact element suffers a loss of accuracy in the results as the number of beam

elements in the model decreases; however, this is not the case for the WRC elements,

which maintain the correct solution. Moreover, it has been shown that if one replaces

the shape functions N1 with H1 and N2 with H2, while G1 and G2 are set to zero, the results


315

from the ANSYS contact element are equal to results from a linear WRC element.

Nonetheless, both systems give similar results for wheel-rail separation.

Subsequently, the author develops both a modal model and finite element model

incorporating unsprung wheels traversing a bridge. The unsprung wheel has the same

vertical position as the point of the beam directly underneath. However, since the

moving unsprung wheel is moving horizontally, its vertical acceleration is not the

same as the vertical acceleration of the point of the beam directly underneath. In fact

the vertical acceleration of the moving unsprung mass is equal to the local vertical

acceleration of the beam plus an additional convective term. A comparison of the final

form of the modal model with the final form of the finite element model reveals that if

the mode shapes of the modal solution are replaced by the finite element weighting

and shape functions, the finite element solution is derived, apart from the stiffness

matrix and the nodal shear force and bending moment.

These unsprung systems were developed to allow comparison with Akin & Mofid’s

(1989) paper, which omits the additional convective acceleration; thus, their solution

is inaccurate by a factor of 4 for a moving unsprung mass traversing a cantilever

beam. Moreover, it has been shown from the results that the unsprung solution is

comparable to the sprung wheel system provided that the wheel-rail contact (WRC)

element is given a reasonable large Hertzian stiffness and is not allowed undergo

separation. It can be seen from the results that the deflection at mid-span of the simply

supported beam subjected to a moving unsprung wheel (modal solution) can be

described accurately with only 3 modes, but at least 12 modes are required for an

accurate description of the shear force. Furthermore, the modal and finite element


316

model differ by only 5% in predicting the internal forces in the beam; and are

remarkable similar to each other.

Another limitation of the ANSYS finite element program was contact elements that

could not model irregularities. Fortunately, the author was able to modify the sprung

(WRC element) and unsprung models (modal and finite element models) to simulate

irregularities along the track. It was found that in order to simulate the irregularity

function in the wheel-rail contact (WRC) element; one must apply forces and

moments to the nodes of the beam element directly beneath the wheel as well as apply

a nodal force to the centre of the wheel. The unsprung systems follow a similar

development; however, the vertical acceleration of the unsprung wheel introduces

additional terms such as a second derivative with respect to x for the irregularity

function. One finds that the results from the developed models compare reasonably

well with the Manchester Benchmark simulation i.e. a twin-axle freight wagon

traversing a track with lateral or vertical irregularities (Track Case 3 or 4).

The train model used extensively throughout this thesis consists of a single 201 Class

locomotive and three Mark3 railway coaches, with an overall length from the front

axle to the rear axle of 85.88 m, traversing the Boyne Viaduct. Examining the results

from the Boyne Viaduct subjected to the train travelling at a constant speed, one finds

that the maximum dynamic amplification factor for the deflection at mid-span of the

bridge (DAFU) tends to be always less than 1.1 for realistic vehicle speeds. This value

comes from modelling the train as a series of multiple moving point forces. Modelling

the train as two-dimensional and three-dimensional railway vehicles tends to lower

the maximum DAFU to a maximum value of 1.07 for realistic vehicle speeds. The


317

results also show that the train modelled with sprung or unsprung wheels gives fairly

similar results. It has been shown in Chapter 3 that resonance of the bridge can occur

if a critical speed is reached. This critical speed can be related to: (1) the traffic speed;

and (2) the repetitive vehicle loading of the bridge. The critical speed of the three-

dimensional Boyne Viaduct is 1627 km/hr, which is an unrealistic speed; however,

several freight wagons with a short repetitive distance between axles, make the Boyne

Viaduct vulnerable to resonance as they transverse the bridge at realistic speeds, as

observed in Section 3.3.3.5

The primary aims of this thesis were to create a contact element that could be used to

simulate a train traversing a railway bridge. From the results presented in this thesis,

the reader should agree that the wheel-rail contact (WRC) element is exceedingly

robust in capturing the dynamic response of both the bridge and vehicle. This model

has the advantage of also simulating wheel-rail separation, rigid rails, braking and

accelerating as well as vertical and lateral irregularities. The unsprung modal and

finite element model may not be as versatile as the WRC element but their results are

comparable with the results from the WRC element. The unsprung models were

developed to address Akin & Mofid’s (1989) paper, which omits the convective

acceleration term. Nevertheless, many authors such as Yang and Wu (2001) are still

comparing their developed models with Akin and Mofid’s (1989) inaccurate solution,

ignoring the issue of convective acceleration (Bowe & Mullarkey, 2008)

Referring to the drawings presented by the author in Figures 6.23, 6.29 and 6.34, the

reader can see the train traversing different track profiles, which is a primary aim of

this thesis to simulate track irregularities.


318

7.2 Recommendations for future work

As this thesis has developed, the author’s dependency on the ANSYS finite element

program has reduced. Originally, it was thought that this versatile finite element

program would satisfy the author’s needs in the study of railway dynamics; however,

it has been shown throughout all the chapters that ANSYS has limitations. Its greatest

asset still remains its time-integration scheme to solve both the numerical and modal

transient analyses. Additionally, its best element became the empty mass, stiffness

and damping matrix element [MATRIX27], which would be used in the development

of the wheel-rail contact element as well as the unsprung modal and finite element

solution. Therefore, it would be highly advantageous for the author to develop a finite

element program with these mathematical implementations independent of ANSYS’s

program and license. Other developments that would be beneficial for future study

are:

• Extending the wheel-rail contact element to include the effects of a train

traversing a curved track or bridge structure.

• Modification of the modal unsprung solution to simulate more complex bridge

structures (only simple beam structures were adopted in this study).

• A more comprehensive study of two trains travelling in opposite directions i.e.

with different train lengths, speeds, weights, axle spacing and arriving times.

• Examining other Irish Rail bridges that have to experimental data recorded so

that one can verify the numerical simulations with the experimental data.

• A re-examination of the wheel-rail contact element algorithm in order to

improve its current time of execution.

Appendix A – Conventions & Elastic beam theory

319

Appendix A

Conventions & Elastic beam theory

A.1 Introduction

This appendix has three main purposes: (1) to establish a consistent sign convention

that is used throughout this thesis, (2) to develop the moment-curvature relationship,

(3) to derive the Bernoulli-Euler differential equation for an elastic beam. In addition,

the equation of motion for a spring, that is, axial motion, is derived.

A.2 Conventions

A.2.1 Convention for coordinate axes

This thesis uses a right-handed Cartesian system of mutually perpendicular axes Ox,

Oy and Oz, similar to that of Coates et al. (1997). By choosing two positive directions

for the x and the y axes, the positive direction for the z-axis is the advance of a right-

handed screw when turned from the x-axis to the y-axis through the right angle

between the positive directions of these two axes. Figure A.1 presents the right-

handed systems used throughout this thesis.


320

90o

o90

xO

zy

O x

z

y

Figure A.1: Right-handed co-ordinate system

A.2.2 Convection for moments

The convention for moment M, of magnitude M, acting on a plane is shown in Figure

A.2a. According to Coates et al. (1997), the moment can be represented by means of a

vector with double-arrow heads whose length corresponds to the magnitude M and

whose direction is normal to the plane in which it acts. The direction of the arrow is

that of the advance of a right-handed screw turned in the same sense as the moment M

(Coates et al., 1997).

A moment vector parallel to a coordinate axis is considered positive if its direction is

the same as the positive direction of that axis as illustrated in Figure A.2b.

O x

z

M

outwards

inwards

(a)

outwards


321

xM

Mz

yM

z

xO

Figure A.2: Convention for positive moments acting (a) in a plane (b) about the

coordinate axes x, y, and z (Coates et al., 1997).

A.2.3 Convections for internal forces and moments of a beam

The x-axis is the longitudinal centroidal axis of the beam, the y and z-axes are

principal axes of the beam’s cross-section with the origin located at the centroid of the

cross section; thus, the products of inertia of the cross-section are zero. If the outward

normal to the beam’s cross-section is in the positive x-coordinate direction, then the

positive axial force (P) acting through the centroid, the positive shear forces (y

Q

andz

Q ) passing through the centroid, the positive torque (T) and positive bending

moments (y

M and z

M ) are in the positive coordinate directions; however, if the

outward normal to the cross-section is in the negative coordinate direction, then the

positive axial force, shear forces and moments on that face are in the negative

coordinate direction in accordance with Newton’s Third Law (action and reaction are

equal and opposite) as shown in Figure A.3.

outwards


322

Figure A3: Positive internal forces and moments acting on the cross-sections of a

three-dimensional beam segment

A.2.4 Convection for stresses

The convention for stresses acting on an infinitesimal parallelepiped is as follows: a

normal stress is defined as positive if it is tensile and the normal stress is defined as

negative if it is compressive. In accordance with Coates et al. (1997), the positive

normal and shear stresses acting on a face are in the positive coordinate directions

when the outward normal to that face is in the positive coordinate directions, while,

the positive normal and shear stresses acting on a face are in the negative coordinate

directions when the outward normal to that face is in the negative coordinate

directions in accordance with Newton’s third law as illustrated in Figure A.4. For

shear stress the first subscript denotes the outward normal to the surface, while the

second subscript denotes the direction of the shear stress.

T

T Mz

yM

My

zM

xz

y

Qz

yQ

P

P

Qz

yQ


323

τzy

zxτ

τzx

zyτ

σz

zσ

yzτ

τyz

τyx

yxτ

σy

yσ

xστxz

τxy τxz

xyτ

z

y

xσx

Figure A.4: Positive normal and shear stresses acting on an infinitesimal

parallelepiped (Coates et al., 1997)

A.3 Moment-Curvature Relationship

Figure A.5 shows an undeformed beam axis and cross-section ab in side view in the

x-y plane. Transverse loading is applied in the y-direction and the cross-section ab

moves to its new location a b′ ′ . The cross section of the beam is perpendicular to the

axis of the undeformed beam and remains perpendicular to the axis of the deformed

beam.

dz

dx

dy


324

zθ

p'

p

a'

b'

b

a

y, v

x, u

(z-axis out)

Figure A.5: Displacement of a beam axis and cross-section in the x-y plane due to

transverse loading

The x-component of the displacement of a typical point p of the cross-section is

( ), , , :u x y z t

( ), , , sin zu x y z t y θ= − (A.1)

where y is the distance of p above the centroidal axis of the beam and z

θ is the angle

between the tangent and the x-axis. Since the angle z

θ is small, it is approximately

equal to sinz

θ or tanz

θ , but tanz

θ is the slope of the deformed beam axis; therefore:

sin tanz z z

v

xθ θ θ

∂= = =

∂ (A.2)

where ( ),v x t is the displacement of the longitudinal axis in the positive y-direction.

Substituting Equation (A.2) into (A.1) gives

vu y

x

∂= −

∂ (A.3)

vu y

x

∂= −

∂


325

x, u

z, w

c

d

d'

c'

p

p'

(y-axis in)

Figure A.6: Displacement of a beam axis and cross-section in the x-z plane

The axis and cross-section of the beam in side view are re-examined in the x-z plane.

Figure A.6 presents the undeformed beam axis and cross-section cd. A transverse

loading is applied in the z-direction and the cross-section cd moves to its new location

c d′ ′ . The x-component of the displacement of a typical point p of the cross-section is

( ), , , :u x y z t

( ), , , sin yu x y z t z θ= (A.4)

where z is the distance of p above the centroidal axis of the beam and y

θ is the angle

between the tangent and the x-axis. As before, since the angle y

θ is considered to be

small, it is approximately equal to siny

θ or tany

θ , which is the slope of the deformed

beam axis so that:

sin tany y y

w

xθ θ θ

∂= = = −

∂ (A.5)

wu z

x

∂= −

∂

yθ


326

where w is the displacement of the longitudinal axis in the positive z-direction.

Substituting Equation (A.5) into (A.4) gives:

wu z

x

∂= −

∂ (A.6)

Combining Equation (A.3) and (A.6) gives the total horizontal displacement u as:

v wu y z

x x

∂ ∂= − −

∂ ∂ (A.7)

The linear strain xx

ε in the beam fibres is:

xx

u

xε

∂=

∂ (A.8)


2 2

2 2xx

v wy z

x xε

∂ ∂= − −

∂ ∂ (A.9)

It is assumed that the elastic beam obeys Hooke’s Law, and that Poisson’s ratio is

zero; therefore, the uniaxial stress-strain equation is:

x xxEσ ε= (A.10)

where E is Young’s modulus of elasticity. Substituting Equations (A.9) into (A.10)

gives:


327

2 2

2 2x

v wEy Ez

x xσ

∂ ∂= − −

∂ ∂ (A.11)

Figure A.7 shows that the positive bending moment acting on a cross-section whose

outward normal is in the positive x-direction can be related to the normal stress x

σ as

follows:

z x

A

M y dAσ= −∫ (A.12a)

y x

A

M z dAσ= ∫ (A.12b)


2 22

2 2z

A A

v wM E y dA E yzdA

x x

∂ ∂= +

∂ ∂∫ ∫ (A.13a)

2 22

2 2y

A A

w vM E z dA E yzdA

x x

∂ ∂= − −

∂ ∂∫ ∫ (A.13b)

(y-axis in)

(z-axis out)

Beam centroidal axis

z

x

dA


y

(b) Beam section(a) Stress distribution

M

x

y

z

σx

+y

xσ

xσ

σx

dAz

M

z

y

+z y

Figure A.7: Stress distribution on an elastic beam segment due to bending moment

Stress and moment

(z-axis out)

(y-axis in)


328

The integrals in Equation (A.13) are section properties that are defined as:

2

z

A

y dA I=∫ (A.14a)

2

y

A

z dA I=∫ (A.14b)

0A

yzdA =∫ (since y and z are principal axes) (A.14c)

where Iz and Iy are the second moment of area of the cross-section of the beam.

Substituting Equation (A.14) into (A.13) gives the moment-curvature equation of the

Bernoulli-Euler beam as:

2

2z z

vM EI

x

∂=

∂ (A.15a)

2

2y y

wM EI

x

∂= −

∂ (A.15b)

Equally, substituting Equation (A.15) into (A.11) gives the relationship between the

normal stress and the bending moment as follows:

yzx

z y

M zM y

I Iσ = − + (A.15c)

A.4 Differential equations governing the transverse deformation of a beam

In Figure A.8, an elastic beam is subjected to a transverse load ( ),yp x t , where m is

the mass per unit length and EIz is the flexural property of the beam.


329

y

z

y, v

x

L

p (x)

x dxEI , m

z

Figure A.8: Elastic beam subjected to a transverse load in the x-y plane

Figure A.9 presents a differential beam segment of length dx subjected to a transverse

load ( ),yp x t acting in the positive y direction. The shear force y

Q and bending

moment z

M acting on this particular segment obey the sign convention described in

Section A.2.3.

y

M

Q

dx

p (x)

M + dxx

Mz z

z

y

x

y, v

zy

yQ

dxQ +

Figure A.9: Free body diagram of a beam segment in the x-y plane

Recalling Equation (A.15a), the constitutive law relating bending moment to

curvature states:

2

2z z

vM EI

x

∂=

∂ (A.16)

( ),yp x t

( ),y

p x t

outward


330

For the segment, the sum of moments about the z-axis on the positive face of the free-

body diagram must be equal to zero (ignoring rotatory inertia):

( )( )2

,0

2

yzz y z

p x t dxMM dx Q dx M

x

∂ + + − + = ∂

(A.17)

Simplifying Equation (A.17) gives the following relationship between the bending

moment z

M and the shear force y

Q :

zy

MQ

x

∂= −

∂ (A.18)


2

2y z

vQ EI

x x

∂ ∂= −

∂ ∂ (A.19)

Next, Newton’s second law is applied in the vertical y-direction:

2

2( , ) 2

y

y y by y

Q v vQ dx p x t dx m dx Q mdx

x t tω

∂ ∂ ∂+ + − − =

∂ ∂ ∂ (A.20)

where by

ω is frequency of damping of the beam in the y-direction, (see Appendix D).

Equation (A.20) reduces to the following:

2

2( , ) 2

y

y by

Q v vp x t m m

x t tω

∂ ∂ ∂+ = +

∂ ∂ ∂ (A.21)



331

2 2 2

2 2 22 ( , )z by y

v v vEI m m p x t

x x t tω

∂ ∂ ∂ ∂+ + =

∂ ∂ ∂ ∂ (A.22)

which is the equation of motion of an elastic beam in the y-direction.

Next the equation of motion of the beam in the x-z plane is derived. Figure A.10

shows a differential beam segment of length dx subjected to a load ( )zp x acting in

the positive z direction. The shear force z

Q and bending moment y

M acting on this

particular beam segment obey the sign convention described in Section A.2.3.

z

Q + dxx

Qz

z

y

z, w

x

z

yyy

M

xdxM +

p (x)

dx

Q

M

Figure A.10: Free body diagram of a beam segment in the x-z plane

Recalling Equation (A.15b), the constitutive law relating bending moment to

curvature states:

2

2y y

wM EI

x

∂= −

∂ (A.23)

The sum of moments about the y-axis on the positive face of the free-body diagram

must be equal to zero (ignoring rotatory inertia):

( ),zp x t

inward


332

( )( )( )

2,

02

y z

y z y

M p x t dxM dx Q dx M

x

∂+ + − + =

∂ (A.24)

Simplifying Equation (A.24) gives the following relationship between the bending

moment y

M and the shear force z

Q :

y

z

MQ

x

∂=

∂ (A.25)


2

2z y

wQ EI

x x

∂ ∂= −

∂ ∂ (A.26)

Next, Newton’s second law is applied in the vertical z-direction:

2

2( , ) 2z

z z bz z

Q w wQ dx p x t dx m dx Q mdx

x t tω

∂ ∂ ∂+ + − − =

∂ ∂ ∂ (A.27)

where bzω is frequency of damping of the beam in the z-direction, (see Appendix D).

Equation (A.27) is simplified as follows:

2

2( , ) 2z

z bz

Q w wp x t m m

x t tω

∂ ∂ ∂+ = +

∂ ∂ ∂ (A.28)

Substituting Equation (A.26) into (A.28):


333

2 2 2

2 2 22 ( , )y bz z

w w wEI m m p x t

x x t tω

∂ ∂ ∂ ∂+ + =

∂ ∂ ∂ ∂ (A.29)

which is the equation of motion of elastic beam in the z-direction.

A.5 Differential equation governing the longitudinal deformation of a beam

In Figure A.11a, an elastic beam is subjected to a longitudinal load ( ), .xp x t The

longitudinal loading causes a cross-section ab to move to a new location a b′ ′ as

shown in Figure 11b. The displacement of a typical point p on the cross section of the

beam is ( ),u x t .

(z-axis out)

u

b'

a'p'p

b

a

y, v

x, u

Figure A.11: (a) Elastic beam subjected to a longitudinal load (b) displacement of a

beam cross-section in the x direction

The linear strain x

ε in the fibres of the beam can be defined as:

(a)

(b)

( ), .xp x t , , E A m

L

, x u

, y v

z dxx


334

xx

u

xε

∂=

∂ (A.30)


x

uE

xσ

∂=

∂ (A.31)

Figure A.12 shows that the positive axial force acting on the positive face of the

segment can be related to the normal stress x

σ acting on the cross section as follows:

x

A

P dAσ= ∫ (A.32)

(z-axis out)

P

(a) Stress distribution (b) Beam section

σx

+y

xσ

z

yy

dA

x


Figure A.12: Stress distribution on an elastic beam segment due to axial force


A

u uP E dA EA

x x

∂ ∂= =

∂ ∂∫ (A.33)

In Figure A.13, a differential beam segment of length dx is subjected to load ( )xp x

acting in the positive x direction.


335

x

P

z

y, v

x, u

P

xdxP +

p (x)

dx

Figure A.13: Beam segment undergoing axial loading

Newton’s second law is applied in the horizontal direction:

( )2

2, 2

x bx

P u uP dx p x t dx P m dx mdx

x t tω

∂ ∂ ∂+ + − − =

∂ ∂ ∂ (A.34)

where bxω is frequency of damping of the beam in the x-direction, (see Appendix D).

Simplifying Equation (A.34) yields:

( )2

2, 2

x bx

P u up x t m m

x t tω

∂ ∂ ∂+ − =

∂ ∂ ∂ (A.35)

Substituting Equation (A.33) into (A.35) and rearranging gives the equation of motion

for axial vibration of an elastic spring:

( )2

22 ,

bx x

u u uEA m m p x t

x x t tω

∂ ∂ ∂ ∂ − + + =

∂ ∂ ∂ ∂ (A.36)

( ),x

p x t

outward


336

A.6 Differential equation governing the torsional deformation of a beam

In this section, the equation of motion for an elastic beam subjected to torque is

derived. It is assumes that every cross section remains plane and perpendicular to the

longitudinal axis. In Figure A.14, the torsional loading causes the cross-section to

rotate about the longitudinal axis. A typical point p moves to its new location .p′ The

rotation of the cross-section is x

θ .

p'x, u

y, v

a

b

p

(z-axis out)

w

v

p'

p

(x-axis out)z

O

Figure A.14: Displacement due to applied torque of a beam cross-section (a) in the

xy plane (b) in the yz plane

Examining the displacement of a typical point p to p′ , assuming the displacement and

rotational angle is small, one gets (ignoring warping):

( ) ( ), , , , sin ,x xv x y z t r z x tθ α θ= − = − (A.37a)

( ) ( ), , , cos ,x xw x y z t r y x tθ α θ= = (A.37b)

where r is the distance between the beam’s longitudinal axis and the point p. The

shear strains in the fibres of the beam can then be defined as:

α

(a)

(b)

xθ

α

r


337

1

2xy

v u

x yε

∂ ∂= +

∂ ∂ (A.38a)

1

2xz

w u

x zε

∂ ∂ = +

∂ ∂ (A.38a)


2

xxy

z

x

θε

∂ = −

∂ (A.39a)

2

xxz

y

x

θε

∂ =

∂ (A.39a)

For a linear-elastic infinitesimal body, Hooke’s Law for shear states that:

2xy xy

Gτ ε= (A.40a)

2xz xz

Gτ ε= (A.40a)

where G is the shear modulus of elasticity. Substituting Equation (A.40) into (A.39)

gives the following expression:

xxy

Gzx

θτ

∂= −

∂ (A.41a)

xxz

Gyx

θτ

∂=

∂ (A.41a)

Examining the shear stress distribution of the cross-section of body in Figure A.15,

one finds that positive moment about the centroidal longitudinal axis i.e. torque, is as

follows:


338

( )xz xy

A

T y z dAτ τ= −∫ (A.42)


( )2 2x

A

T G y z dAx

θ∂= +

∂ ∫ (A.43)

+z

+y dA

T

r

τxz

xyτ

zdA

z

(b) Cross-section(a) Shear stress distribution

Figure A.15: Shear stress distribution on the cross-section of an elastic body

The integral in Equation (A.43) is defined as:

( )2 2 2

p

A A

y z dA r dA I+ = =∫ ∫ (A.44)

where p

I is the polar moment of inertia of the cross-section of the elastic body.

Substituting Equation (A.44) into (A.43) gives the relationship between the torque

and angle of twist of the body as:

xp

T GIx

θ∂=

∂ (A.45)


339

The inertia of a parallelepiped of a beam undergoing a positive rotation about the

centroidal longitudinal axis as shown in Figure A.16 is as follows:

( )2

2

m vdydz dx

A t

∂

∂ (in the y-direction) (A.46a)

( )2

2

m wdydz dx

A t

∂

∂ (in the z-direction) (A.46b)

where m is the mass per unit length of the beam.

Figure A.16: Segment of a beam subjected to a torque ( ),xm x t

The acceleration terms with the aid of Equation (A.37) are:

22

2 2

xvz

t t

θ∂∂= −

∂ ∂ (A.47a)

22

2 2

xwy

t t

θ∂∂=

∂ ∂ (A.47a)

TT dx

x

∂+

∂

( ),x

m x t

outward

dx

,x u

,y v

,z w

T


340

Equation (A.46a) is multiplied by z− and Equation (A.46b) by y+ , which gives:

( )2

2

m vz dydz

A t

∂−

∂ (A.48a)

and

( )2

2

m wy dydz

A t

∂

∂ (A.48b)

Substituting Equations (A.47) into Equation (A.48) and then adding Equations

(A.48a) to (A.48b) and integrating over the cross-section gives the rotational inertia

about the longitudinal axis as:

( )2 2

2 2

2 2

x x

p

A

m my z dA I

A t A t

θ θ∂ ∂+ =

∂ ∂∫ (A.49)

Newton’s second law tell us that the rotational inertia about the longitudinal axis of

the beam segment of Figure A.16 is equal to the sum of the moments about the x-axis:

( ) ( )2

2, 2 ,

p x x x

x p brx x

mI Tm x t GI m m x t

A t x x x t

θ θ θω

∂ ∂ ∂∂ ∂ = + = + +

∂ ∂ ∂ ∂ ∂

or

( )2

22 ,

px x x

p brx x

mIGI m m x t

x x A t t

θ θ θω

∂ ∂ ∂∂ − + + =

∂ ∂ ∂ ∂ (A.50)

where ( ),xm x t is the torque moment about the x-axis and brx

ω is the frequency of

damping of the beam about the x-axis.

Appendix B – FEM applied to equations of motion of a beam

341

Appendix B

FEM applied to equations of motion of a beam

B.1 Introduction

In this appendix, the finite element method is applied to the differential equations

governing the motion for the beam. The Galerkin weighted residual method is used.

B.2 Beam Element in the x-y plane

In Figure B.1, a beam finite element of length l is subjected to a uniformly distributed

load, ( ),yp x t acting in the positive y direction. The nodal displacements, forces and

moments have the following convention: the nodal shear forces, 1ˆ

yQ and 2ˆ

yQ , and

nodal displacements 1v and 2v are positive in the positive y-direction and the nodal

bending moments , 1ˆ

zM and 2

ˆz

M , and the nodal rotations, 1ˆz

θ and 2ˆz

θ , are positive in

the positive direction of the z-axis. The integer subscripts denote the local node of the

element.


342

l

21 x

z

Figure B.1: Beam element subjected to a uniformly distributed load in x-y plane.

Figure B.2 presents the sign convention for internal forces, i.e. positive axial force,

positive shear force and positive bending moment used in the elastic beam theory in

Appendix A. Comparing Figure B.1 with B.2, it is found that:

( ) ( )1 2ˆ ˆ0 ; y y y yQ Q Q Q l= − = + (B.1a)

( ) ( )1 2ˆ ˆ0 ; z z z zM M M M l= − = + (B.1b)

y y

z zP P

Q

M

Q

M

Figure B.2: Sign convection for positive internal forces and bending moments

The beam element in the x-y plane, has four degrees of freedom; therefore, a cubic

displacement function ( ),v x t with four coefficients ( )1, 4ia i = is used:

2 3

1 2 3 4v a a x a x a x= + + + (B.2)

z

x

y, v

0

1 1ˆˆ ,z zM θ 2 2

ˆˆ ,z zM θ

1 1ˆ ,yQ v

2 2ˆ ,yQ v

, zEI m

( ),yp x t


343

Using Equation (A.2), the slope at each node is defined as:

2

2 3 42 3z

dva a x a x

dxθ = = + + (B.3)

The coefficients ( )1, 4ia i = used are evaluated from the nodal value of displacement

and its first derivate, which are evident from inspection of Figure A.5 and Figure B.1 as:

1 1ˆ and

z z

dvv v

dxθ θ= = = at x = 0 (B.4a)

2 2ˆ and

z z

dvv v

dxθ θ= = = at x = l (B.4b)

Substituting Equation (B.4) into Equation (B.2) and (B.3), one obtains in matrix form:

1 1

1 2

2 3

2 3

2

42

1 0 0 0

ˆ 0 1 0 0

1

ˆ 0 1 2 3

z

z

v a

a

v al l l

al l

θ

θ

=

(B.5)

Inverting Equation (B.5) gives

11

2 1

2 23 2

4 23 2 3 2

1 0 0 0

0 1 0 0ˆ

3 2 3 1

ˆ2 1 2 1

z

z

va

a

a vl l l l

a

l l l l

θ

θ

= − − − −

(B.6)

Substituting Equation (B.6) into (B.2) and gathering terms yields

2 3 2 3 2 3 2 3

1 1 2 22 3 2 2 3 2

3 2 2 3 2ˆ ˆ1z z

x x x x x x x xv v x v

l l l l l l l lθ θ

= − + + − + + − + − +


344

( ) ( ) ( ) ( )1 1 1 1 2 2 2 2ˆ ˆz zN x v G x N x v G xθ θ= + + + (B.7)

where ( )1N x , ( )1G x , ( )2N x and ( )2G x are known as shape functions and are

defined as:

( )2 3

1 2 3

3 21

x xN x

l l= − + , ( )

2

1 2 3

6 6x xN x

l l′ = − + , ( )1 2 3

6 12xN x

l l′′ = − + , ( )1 3

12N x

l′′′ =

( )2 3

1 2

2x xG x x

l l= − + , ( )

2

1 2

4 31

x xG x

l l′ = − + ( )1 2

4 6xG x

l l′′ = − + , ( )1 2

6G x

l′′′ =

( )2 3

2 2 3

3 2x xN x

l l= − , ( )

2

2 2 3

6 6x xN x

l l′ = − , ( )2 2 3

6 12xN x

l l′′ = − , ( )2 3

12N x

l′′′ = −

( )2 3

2 2

x xG x

l l= − + , ( )

2

2 2

2 3x xG x

l l′ = − + , ( )2 2

2 6xG x

l l′′ = − + , ( )2 2

6G x

l′′′ = (B.8)

Each dash over the shape functions ( )1N x ( )1G x , ( )2N x and ( )2G x represents a

derivative with respect to x. A plot of the beam element shape functions can also be

seen in Figure B.3.

l

x

1

0

l

x

1

0

1

x

l0

1

x

l

0

Figure B.3: Beam element shape functions

N1

G1

N2

G2

45o

45otan 45 1o =


345

Recalling Equation (A.22) with z

EI constant and omitting the damping term, the

differential equation of motion for an elastic beam is given as:

( )4 2

4 2,

z y

v vEI m p x t

x t

∂ ∂+ =

∂ ∂ (B.9)

where E is Young’s modulus of elasticity, Iz is the moment of inertia and m is the

mass per unit length. It should be noted that m Aρ= , whereby A is the cross-sectional

area of the beam and ρ is density of the beam material.

In order to apply the method of weighted residuals, one multiplies both sides of

Equation (B.9) by a weighting function vδ and integrates along the element length:

( )4 2

4 2

0

, 0

l

z y

v vEI m p x t vdx

x tδ

∂ ∂+ − =

∂ ∂ ∫ (B.10)

The term 4

4

0

l

z

vEI v dx

xδ

∂

∂∫ in Equation (B.10) is now integrated by parts twice using the

following general expression:

0

0 0

l ll

udv uv vdu= −∫ ∫ (B.11)

resulting in the following:

4 3 3

4 3 3

0 00

ll l

z z z

v v v vEI v dx vEI EI dx

x x x x

δδ δ

∂ ∂ ∂ ∂= −

∂ ∂ ∂ ∂∫ ∫

3 2 2 2

3 2 2 2

00 0

l l l

z z z

v v v v vvEI EI EI dx

x x x x x

δ δδ

∂ ∂ ∂ ∂ ∂= − +

∂ ∂ ∂ ∂ ∂∫


346

3 3 2 2 2 2

3 3 2 2 2 2

000

l

z z z z z

l l

v v v v v v v vvEI vEI EI EI EI dx

x x x x x x x x

δ δ δδ δ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂= − − + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂∫

2 2

2 20 0 0

l

y y z z zll

v v v vvQ vQ M M EI dx

x x x x

δ δ δδ δ

∂ ∂ ∂ ∂= − + − + +

∂ ∂ ∂ ∂∫

2 2

2 1 2 1 2 200 0

ˆ ˆ ˆ ˆl

y y z z zll

v v v vv Q v Q M M EI dx

x x x x

δ δ δδ δ

∂ ∂ ∂ ∂= − − − − +

∂ ∂ ∂ ∂∫ (B.12)

where the shear force y

Q and the bending moment z

M are defined by Equations

(A.19) and (A.16), respectively. Substituting Equation (B.12) into (B.10) and

rearranging gives:

2 2 2

2 2 2

0

l

z

v v vEI m v dx

x x t

δδ

∂ ∂ ∂+

∂ ∂ ∂ ∫

( ) 1 1 2 2000

ˆ ˆˆ ˆ,

l

y y z y zll

v vp x t vdx v Q M v Q M

x x

δ δδ δ δ

∂ ∂= + + + +

∂ ∂∫ (B.13)

We now apply Galerkin’s method of weighted residuals, where the four shape

functions are used as weighting functions. The result is the following four equations:

( ) ( ) ( ) ( ) ( )2 2

1 1 1 1 1 1 1 1 2 1 22 2

0 0

ˆ ˆˆ ˆ, 0 0

l l

z y y z y z

v vEI N mN dx p x t N dx N Q N M N l Q N l M

x t

∂ ∂′′ ′ ′+ = + + + + ∂ ∂

∫ ∫ (B.14a)

( ) ( ) ( ) ( ) ( )2 2

1 1 1 1 1 1 1 1 2 1 22 2

0 0

ˆ ˆˆ ˆ, 0 0

l l

z y y z y z

v vEI G mG dx p x t Gdx G Q G M G l Q G l M

x t

∂ ∂′′ ′ ′+ = + + + + ∂ ∂

∫ ∫ (B.14b)

( ) ( ) ( ) ( ) ( )2 2

2 2 2 2 1 2 1 2 2 2 22 2

0 0

ˆ ˆˆ ˆ, 0 0

l l

z y y z y z

v vEI N mN dx p x t N dx N Q N M N l Q N l M

x t

∂ ∂′′ ′ ′+ = + + + + ∂ ∂

∫ ∫ (B.14c)

( ) ( ) ( ) ( ) ( )2 2

2 2 2 2 1 2 1 2 2 2 22 2

0 0

ˆ ˆˆ ˆ, 0 0

l l

z y y z y z

v vEI G mG dx p x t G dx G Q G M G l Q G l M

x t

∂ ∂′′ ′ ′+ = + + + + ∂ ∂

∫ ∫ (B.14d)

where each dash on the displacement function indicates a derivative with respect to x.


347

Rewriting Equation (B.14) in matrix form gives:

1 1

2 21 1

2 2

2 20

2 2

l

z

N N

G Gv vEI m dx

N Nx t

G G

′′ ′′ ∂ ∂ + ′′ ∂ ∂ ′′

∫

( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

1 1 1 11

1 1 1 11

1 1 2 2

2 2 2 220

2 2 2 22

0 0

0 0ˆ ˆˆ ˆ,

0 0

0 0

l

y y z y z

N N N l N lN

G G G l G lGp x t dx Q M Q M

N N N l N lN

G G G l G lG

′ ′ ′ ′

= + + + + ′ ′

′ ′

∫ (B.15)

Next one substitutes Equation (B.7) in (B.15) giving:

[ ] [ ]

111 1

1 1 1 1

1 1 2 2 1 1 2 2

2 2 2 20 0

2 22 2

ˆ ˆ

ˆ ˆ

l l

z z

z

z z

vvN N

G GEI N G N G dx m N G N G dx

N v N v

G G

θ θ

θ θ

′′ ′′

′′ ′′ ′′ ′′ + ′′

′′

∫ ∫

&&

&&

&&

&&

( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

1 1 1 11

1 1 1 11

1 1 2 2

2 2 2 220

2 2 2 22

0 0

0 0ˆ ˆˆ ˆ,

0 0

0 0

l

y y z y z

N N N l N lN


N N N l N lN

G G G l G lG

′ ′ ′ ′

= + + + + ′ ′

′ ′

∫ (B.16)

where each dot over a nodal displacement indicates a derivative with respect to t. The

shape functions and their derivatives are evaluated at 0x = and x l= using Equation

(B.8). Substituting the results into the right hand side of Equation (B.16) leads to the

following:

[ ] [ ]

111 1

1 1 1 1

1 1 2 2 1 1 2 2

2 2 2 20 0

2 22 2

ˆ ˆ

ˆ ˆ

l l

z z

z

z z

vvN N


N v N v

G G

θ θ

θ θ

′′ ′′

′′ ′′ ′′ ′′ + ′′

′′

∫ ∫

&&

&&

&&

&&


348

( )

1

1

1 1 2 2

20

2

1 0 0 0

0 1 0 0ˆ ˆˆ ˆ,0 0 1 0

0 0 0 1

l

y y z y z

N

Gp x t dx Q M Q M

N

G

= + + + +

∫ (B.17)

Multiplying the column vector and the row vector of shape functions in Equation

(B.17) gives:

11 1 1 1 1 2 1 2 1 1 1 1 1 2 1 2

1 1 1 1 1 2 1 2 1 1

2 1 2 1 2 2 2 2 20 0

2 1 2 1 2 2 2 2 2

ˆ

ˆ

y

l l

z

z

y

z

vN N N G N N N G N N N G N N N G

G N GG G N G G G NEI dx m

N N N G N N N G v

G N G G G N G G

θ

θ

′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ +

′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′

∫ ∫

1

1 1 1 1 2 1 2 1

2 1 2 1 2 2 2 2 2

2 1 2 1 2 2 2 22

ˆ

ˆ

y

z

y

z

v

GG G N G Gdx

N N N G N N N G v

G N G G G N G G

θ

θ

&&

&&

&&

&&

( )

11

1 1

20 2

22

ˆ

ˆ,

ˆ

ˆ

y

l

z

y

y

z

QN

G Mp x t dx

N Q

G M

= +

∫ (B.18)

Carrying out the integration on the left hand side of Equation (B.18) gives the

following matrix equation governing transverse motion of the beam element in the x-y

plane as:

11

2 2 2 2

1 1

3

2 2

2 2 2 2

2 2

12 6 12 6 156 22 54 13

ˆ ˆ6 4 6 2 22 4 13 3

12 6 12 6 54 13 156 22420

ˆ6 2 6 4 13 3 22 4 ˆ

z zz

z z

vvl l l l

l l l l l l l lEI ml

v vl l l ll

l l l l l l l l

θ θ

θ θ

− − − − + − − − − − − − −

&&

&&

&&

&&

( )

11

1 1

20 2

22

ˆ

ˆ,

ˆ

ˆ

y

l

z

y

y

z

QN

G Mp x t dx

N Q

G M

= +

∫ (B.19)


349

The first term on the right-hand side of Equation (B.19) can be written as:

( ) ( )

2 3

2 3

2 31

2 31

2 320 0

2 3

2

2 3

2 3

3 21

2

, ,3 2

l l

y y

x x

l lN x x

xG l l

p x t dx p x t dxN x x

G l l

x x

l l

− +

− + =

−

− +

∫ ∫ (B.20)

Equation (B.20) is evaluated easily when ( ) ( ),y yp x t p t= i.e. independent of x:

( ) ( ) ( )

3 4

2 3

0

2 3 4

1

2 3 2

01

3 420

22 3

20

3 4

2 3

0

2

2 0.5000

2 3 4 0.0833,

0.5000

0.08332

3 4

l

l

l

y y yl

l

x xx

l l

x x xN l

l lG lp x t dx p t p t

N lx xG ll l

x x

l l

− +

− +

= =

− −

− +

∫ (B.21)


350

B.3 Beam Element in the x-z plane

A beam finite element of length l is subjected to a uniformly distributed load, ( ),zp x t

acting in the positive z direction as illustrated in Figure B.4. The coordinate system,

defined by (x, z) for the beam element, is such that the nodal shear forces, 1ˆ

zQ and

2ˆ

zQ , nodal displacement 1w and 2w are positive in the positive z-direction and the

nodal bending moment, 1ˆ

yM and 2ˆ

yM , and the nodal rotation, 1ˆ

yθ and 2ˆ

yθ , are

positive in the positive direction of the y-axis.

l

y

x1 2

Figure B.4: Beam element subjected to a uniformly distributed load in x-z plane.

The sign convention for positive axial force, shear force and bending moment

described in Section A.2.3 can be seen in Figure B.5.

z z

y y

P PQ

M

Q

M

Figure B.5: Sign convection for positive internal forces and bending moments

1 1ˆˆ ,y yM θ

2 2ˆˆ ,y yM θ

1 1ˆ ,zQ w 2 2

ˆ ,zQ w

, yEI m

( ),zp x t

y

x

z, w

0


351

Comparing Figure B.4 with B.5, it can be seen that:

( ) ( )1 2ˆ ˆ0 ; z z z zQ Q Q Q l= − = + (B.22a)

( ) ( )1 2ˆ ˆ0 ; y y y yM M M M l= − = + (B.22b)

The beam element in the x-z plane, also has four degrees of freedom; therefore, a

cubic displacement function with four coefficients is used ( )5,8ia i = , as follows:

2 3

5 6 7 8w a a x a x a x= + + + (B.23)

Using Equation (A.5), the slope at each node is defined as:

2

6 7 82 3y

dwa a x a x

dxθ− = = + + (B.24)

The coefficients ( )5,8ia i = used are evaluated from the nodal values of displacement

and its first derivates, which is evident from an inspection of Figure A.6 and Figure

B.4 as:

1 1ˆ and

y y

dww w

dxθ θ= − = = − at x = 0 (B.25a)

2 2ˆ and

y y

dww w

dxθ θ= − = = − at x = l (B.25b)

Substituting Equation (B.25) into Equation (B.23) and (B.24), one obtains in matrix

form:


352

15

1 6

2 32 7

2

82

1 0 0 0

ˆ 0 1 0 0

1

ˆ 0 1 2 3

y

y

w a

a

w al l l

al l

θ

θ

− =

− − −

(B.26)


15

16

2 227

8 23 2 3 2

1 0 0 0

0 1 0 0ˆ

3 2 3 1

ˆ2 1 2 1

y

y

wa

a

wa l l l l

a

l l l l

θ

θ

−

= − − − −

(B.27)


2 3 2 3 2 3 2 3

1 1 2 22 3 2 2 3 2

3 2 2 3 2ˆ ˆ1 y y

x x x x x x x xw w x w

l l l l l l l lθ θ

= − + + − + − + − + −

or

( ) ( ) ( ) ( )1 1 1 1 2 2 2 2ˆ ˆ

y yw N x w G x N x w G xθ θ= − + − (B.28)

where ( )1N x , ( )1G x , ( )2N x and ( )2G x are the shape functions and are defined in

Equation (B.8). Recalling Equation (A.29) with y

EI constant and omitting the

damping term, the differential equation of motion for an elastic beam is given as:

( )4 2

4 2,

y z

w wEI m p x t

x t

∂ ∂+ =

∂ ∂ (B.29)


353

where E is Young’s modulus of elasticity, Iy is the moment of inertia and m is the

mass per unit length.


Equation (B.29) by a weighting function wδ and integrates along the element length:

( )4 2

4 2

0

, 0

l

y z

w wEI m p x t wdx

x tδ

∂ ∂+ − =

∂ ∂ ∫ (B.30)

The term 4

4

0

l

y

wEI w dx

xδ

∂

∂∫ in Equation (B.30) is now integrated by parts twice using

Equation (B.11), resulting in the following:

4 3 3

4 3 3

0 00

ll l

y y y

w w w wEI w dx wEI EI dx

x x x x

δδ δ

∂ ∂ ∂ ∂= −

∂ ∂ ∂ ∂∫ ∫

3 2 2 2

3 2 2 2

00 0

l l l

y y y

w w w w wwEI EI EI dx

x x x x x

δ δδ

∂ ∂ ∂ ∂ ∂= − +

∂ ∂ ∂ ∂ ∂∫

3 3 2 2 2 2

3 3 2 2 2 2

000

l

y y y y y

l l

w w w w w w w wwEI wEI EI EI EI dx

x x x x x x x x

δ δ δδ δ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂= − − + +

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂∫

2 2

2 20

0 0

l

z z y y yll

w w w wwQ wQ M M EI dx

x x x x

δ δ δδ δ

∂ ∂ ∂ ∂= − + + − +

∂ ∂ ∂ ∂∫

2 2

2 1 2 1 2 200 0

ˆ ˆ ˆ ˆl

z z y y yll

w w w ww Q w Q M M EI dx

x x x x

δ δ δδ δ

∂ ∂ ∂ ∂= − − + + +

∂ ∂ ∂ ∂∫ (B.31)

where the shear force z

Q and the bending moment y

M are defined by Equations

(A.26) and (A.23), respectively. Substituting Equation (B.31) into (B.30) and

rearranging gives:


354

2 2 2

2 2 2

0

l

y

w w wEI m w dx

x x t

δδ

∂ ∂ ∂+

∂ ∂ ∂ ∫

( ) 1 1 2 2000

ˆ ˆˆ ˆ,

l

z z y z yll

w wp x t wdx w Q M w Q M

x x

δ δδ δ δ

∂ ∂= + − + −

∂ ∂∫ (B.32)

We now apply Galerkin’s method of weighted residuals, where the four shape

functions are used as weighting functions. The result is the following four equations:

( ) ( ) ( ) ( ) ( )2 2

1 1 1 1 1 1 1 1 2 1 22 2

0 0

ˆ ˆˆ ˆ, 0 0

l l

y z z y z y

w wEI N mN dx p x t Ndx N Q N M N l Q N l M

x t

∂ ∂′′ ′ ′+ = + − + − ∂ ∂

∫ ∫ (B.33a)

( ) ( ) ( ) ( ) ( )2 2

1 1 1 1 1 1 1 1 2 1 22 2

0 0

ˆ ˆˆ ˆ, 0 0

l l

y z z y z y

w wEI G mG dx p x t Gdx G Q G M G l Q G l M

x t

∂ ∂′′ ′ ′− + =− − + − + ∂ ∂

∫ ∫ (B.33b)

( ) ( ) ( ) ( ) ( )2 2

2 2 2 2 1 2 1 2 2 2 22 2

0 0

ˆ ˆˆ ˆ, 0 0

l l

y z z y z y

w wEI N mN dx p x t N dx N Q N M N l Q N l M

x t

∂ ∂′′ ′ ′+ = + − + −

∂ ∂ ∫ ∫ (B.33c)

( ) ( ) ( ) ( ) ( )2 2

2 2 2 2 1 2 1 2 2 2 22 2

0 0

ˆ ˆˆ ˆ, 0 0

l l

y z z y z y

w wEI G mG dx p x t Gdx G Q G M G l Q G l M

x t

∂ ∂′′ ′ ′− + =− − + − +

∂ ∂ ∫ ∫ (B.33d)

where each dash on the displacement function indicates a derivative with respect to x.

It should be noted that Equation (B.33b) and (B.33d) have both been multiplied by -1,

so that the sign of the weighting function corresponds with the sign of the shape

functions in Equation (B.28). Rewriting Equation (B.33) in matrix form, one gets:

1 1

2 21 1

2 2

2 20

2 2

l

y

N N

G Gw wEI m dx

N Nx t

G G

′′ ′′− −∂ ∂ + ′′ ∂ ∂ ′′− −

∫


355

( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

1 1 1 11

1 1 1 11

1 1 2 2

2 2 2 220

2 2 2 22

0 0

0 0ˆ ˆˆ ˆ,

0 0

0 0

l

z z y z y

N N N l N lN


N N N l N lN

G G G l G lG

′ ′ ′ ′− − − −−

= + − + − ′ ′

′ ′− − − −−

∫ (B.34)

Next, one substitutes Equation (B.28) into (B.34) giving:

[ ] [ ]

111 1

1 11 1

1 1 2 2 1 1 2 2

2 22 20 0

2 22 2

ˆ ˆ

ˆ ˆ

l ly y

y

y y

wwN N


w wN N

G G

θ θ

θ θ

′′ ′′ − − ′′ ′′ ′′ ′′− − + − −

′′ ′′− −

∫ ∫

&&

&&

&&

&&

( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

( )( )( )( )

1 1 1 11

1 1 1 11

1 1 2 2

2 2 2 220

2 2 2 22

0 0

0 0ˆ ˆˆ ˆ,

0 0

0 0

l

z z y z y

N N N l N lN


N N N l N lN

G G G l G lG

′ ′ ′ ′− − − −−

= + − + − ′ ′

′ ′− − − −−

∫ (B.35)

where each dot over a nodal displacement indicates a derivative with respect to t. The

shape functions and their derivatives are evaluated at x = 0 and x = l using Equation

(B.8). Substituting the results into the right hand side of Equation (B.35) leads to the

following:

[ ] [ ]

111 1

1 11 1

1 1 2 2 1 1 2 2

2 22 20 0

2 22 2

ˆ ˆ

ˆ ˆ

l ly y

y

y y

wwN N


w wN N

G G

θ θ

θ θ

′′ ′′ − − ′′ ′′ ′′ ′′− − + − −

′′ ′′− −

∫ ∫

&&

&&

&&

&&

( )

1

1

1 1 2 2

20

2

1 0 0 0

0 1 0 0ˆ ˆˆ ˆ,0 0 1 0

0 0 0 1

l

z z y z y

N

Gp x t dx Q M Q M

N

G

− −

= + − + − − −

∫ (B.36)


356


(B.36) gives:

11 1 1 1 1 2 1 2

11 1 1 1 1 2 1 2

22 1 2 1 2 2 2 20

2 1 2 1 2 2 2 2 2

ˆ

ˆ

ly

y

y

wN N N G N N N G

G N G G G N G GEI dx

wN N N G N N N G

G N G G G N G G

θ

θ

′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′− − ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′− −

′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′ − − ′′ ′′ ′′ ′′ ′′ ′′ ′′ ′′− −

∫

( )

1 11 1 1 1 1 2 1 2 1

1 11 1 1 1 1 2 1 2 1

22 1 2 1 2 2 2 2 20 0 2

2 1 2 1 2 2 2 2 222

ˆ

ˆ ˆ,

ˆ

ˆ ˆ

z

l ly y

z

z

yy

w QN N N G N N N G N

MG N GG G N GG Gm dx p x t dx

wN N N G N N N G N Q

G N G G G N G G G M

θ

θ

− − − − − + = + − − − − −

∫ ∫

&&

&&

&&

&&

(B.37)

Integrating the left hand side of Equation (B.37) gives the following matrix equation

governing transverse motion of the beam in the x-z plane:

11

2 2 2 21 1

32 2

2 2 2 2

2 2

12 6 12 6 156 22 54 13

ˆ ˆ6 4 6 2 22 4 13 3

12 6 12 6 54 13 156 22420

ˆ6 2 6 4 13 3 22 4 ˆ

y yy

y y

wwl l l l

EI l l l l l l l lml

w wl l l ll

l l l l l l l l

θ θ

θ θ

− − − − − − − − + − − − −

&&

&&

&&

&&

( )

11

11

20 2

22

ˆ

ˆ,

ˆ

ˆ

z

ly

z

z

y

QN

MGp x t dx

N Q

G M

− = +

−

∫ (B.38)

The first term on the right-hand side of Equation (B.38) can be written as:


357

( ) ( )

2 3

2 3

2 3

1

2 3

1

2 320 0

2 3

2

2 3

2 3

3 21

2

, ,3 2

l l

z z

x x

l l

N x xx

l lGp x t dx p x t dx

N x x

G l l

x x

l l

− +

− − + − =

− −

− − +

∫ ∫ (B.39)

Evaluating Equation (B.39) when ( ) ( ),z zp x t p t= , i.e. independent of x gives the

following:

( ) ( )

3 4

2 3

0

2 3 4

1

2 3 2

01

3 420

22 3

20

3 4

2 3

0

2

2 0.5000

2 3 4 0.0833,

0.5000

0.08332

3 4

l

l

l

z z zl

l

x xx

l l

x x xN l

l lG lp x t dx p t p

N lx xG ll l

x x

l l

− +

− − + − − = =

− −

− − +

∫ ( )t (B.40)

B.4 Beam Element along x-axis

Next one considers the simpler truss or axial element. The element consists of two

pin-connected nodes that can only experience axial deformation. The element is

assumed to behave in a linear-elastic manner. In Figure B.6, the truss element is

subjected to an axial force ( ),x

p x t along the x-axis of the element. The coordinate

system, defined by (x, y) for the axial element, is such that the nodal axial forces, 1P

and 2P , and nodal displacements 1u and 2u are in the positive x-direction.


358

E, A, m

l

z

x, up(x)

y, v

Figure B.6: Axial element subjected to an axial force p(x,t) along its length

Comparing Figure B.6 with Figure A.13, it is seen that:

( ) ( )1 2ˆ ˆ0 ; P P P P l= − = + (B.41)

Since the axial element has only two degrees of freedom, a linear displacement

function with two coefficients ( )9,10ia i = is used as follows:

9 10u a a x= + (B.42)

where the coefficients ( )9,10ia i = used are evaluated from the nodal values of the

displacement:

1u u= at x = 0 (B.43a)

2u u= at x = l (B.43b)

Substituting Equation (B.43) into Equation (B.42), one obtains in matrix form:

91

102

1 0

1

au

au l

=

(B.44)

1 1ˆ , P u

2 2ˆ , P u

( ),x

p x t


359


9 1

10 2

1 0

1 1a u

a ul l

= −

(B.45)


1 21x x

u u ul l

= − +

(B.46a)

or

( ) ( )1 1 2 2u H x u H x u= + (B.46b)

where

( )1 1x

H xl

= − ( )1

1H x

l′ = −

( )2

xH x

l= ( )2

1H x

l′ = (B.47)

A plot of these shape functions of the axial element can be seen in Figure B.7.

l

1

0

x

l

0

1

x

Figure B.7: Plot of the displacement functions for the axial element

H1 H2


360

Recalling Equation (A.36) with EA constant and omitting the damping term, the

differential equation of motion for an axial element is given as:

( )2 2

2 2,

x

u uEA m p x t

x t

∂ ∂− + =

∂ ∂ (B.48)

where E is Young’s modulus of elasticity, A is the cross-sectional area of the beam

and m is the mass per unit length.


Equation (B.48) by a weighting function uδ and integrates along the element length:

( )2 2

2 2

0

, 0

l

x

u uEA m p x t udx

x tδ

∂ ∂− + − =

∂ ∂ ∫ (B.49)

The term 2

2

0

lu

EA u dxx

δ∂

∂∫ in Equation (B.49) is now integrated by parts using Equation

(B.11) resulting in the following:

2

2

00 0

ll lu u u u

EA u dx uEA EA dxx x x x

δδ δ

∂ ∂ ∂ ∂= −

∂ ∂ ∂ ∂∫ ∫

00

l

l

u uuP uP EA dx

x x

δδ δ

∂ ∂= − −

∂ ∂∫

2 10

0

ˆ ˆl

l

u uu P u P EA dx

x x

δδ δ

∂ ∂= + −

∂ ∂∫ (B.50)

where the axial force P is defined in Equations (A.33). Substituting Equation (B.50)

into (B.49) and rearranging gives:


361

( )2

1 22 0

0 0

ˆ ˆ,

l l

x l

u u uEA dx m u dx p x t udx u P u P

x x t

δδ δ δ δ

∂ ∂ ∂+ = + +

∂ ∂ ∂ ∫ ∫ (B.51)

We now apply Galerkin’s method of weighted residuals, where the two shape

functions are used as weighting functions. The result is the following two equations:

( ) ( ) ( )2

1 1 1 1 1 1 22

0 0

ˆ ˆ, 0

l l

x

u uEAH mH dx p x t H dx H P H l P

x t

∂ ∂′ + = + + ∂ ∂

∫ ∫ (B.52a)

( ) ( ) ( )2

2 2 2 2 1 2 22

0 0

ˆ ˆ, 0

l l

x

u uEAH mH dx p x t H dx H P H l P

x t

∂ ∂′ + = + +

∂ ∂ ∫ ∫ (B.52b)

Rewriting Equation (B.52) in matrix form, one gets:

( )( )( )

( )( )

21 11 1 1

1 22

2 22 2 20 01

0ˆ ˆ,

0

l l

x

H H lH H Hu uEA m dx p x t dx P P

H H lH H Hx t

′ ∂ ∂ + = + +

′ ∂ ∂ ∫ ∫ (B.53)

Substituting Equation (B.46b) into (B.53) gives:

[ ] [ ]1 1 1 1

1 2 1 2

2 2 2 20 0

l lH u H uEA H H dx m H H dx

H u H u

′ ′ ′ +

′ ∫ ∫

&&

&&

( )( )( )

( )( )

1 11

1 2

2 220

0ˆ ˆ,

0

l

x

H H lHp x t dx P P

H H lH

= + +

∫ (B.54)

The shape functions are evaluated at x = 0 and x = l using Equation (B.47).

Substituting the results into the right hand side of Equation (B.54) gives the

following:


362

[ ] [ ]1 1 1 1

1 2 1 2

2 2 2 20 0

l lH u H uEA H H dx m H H dx

H u H u

′ ′ ′ +

′ ∫ ∫

&&

&&

( )1

1 2

20

1 0ˆ ˆ,

0 1

l

x

Hp x t dx P P

H

= + +

∫ (B.55)


(B.55) gives:

1 1 1 2 1 1 1 1 2 1

2 1 2 2 2 2 1 2 2 20 0

l lH H H H u H H H H uEA dx m dx

H H H H u H H H H u

′ ′ ′ ′ + ′ ′ ′ ′

∫ ∫&&

&&

( )1 1

20 2

ˆ,

ˆ

l

x

H Pp x t dx

H P

= +

∫ (B.56)

Integrating the left hand side of Equation (B.56) gives the following matrix equation

governing longitudinal motion of a beam element as:

1 1

2 2

1 1 2 1

1 1 1 26

u uEA ml

u ul

− + −

&&

&&( )1 1

20 2

ˆ,

ˆ

l

x

H Pp x t dx

H P

= + ∫ (B.57)

B.5 Torsion Element along the x-axis

To finish, the author examines the torsion effects on an element. Figure B.8 presents a

torsion element subjected to an applied torque ( ),x

p x t about the x-axis of the

element. The coordinate system, defined by (x, y) for the torsion element, is such that

the nodal torques, 1T and 2T , and nodal rotation 1ˆx

θ and 2ˆx

θ are positive in the

direction of the positive x-axis.


363

E, A, m

l

z

x, up(x)

y, v

Figure B.8: Torsion element subjected to a torque force along its length

Comparing Figure B.8 with Figure A.3, it is found that:

( ) ( )1 2ˆ ˆ0 ; T T T T l= − = + (B.58)

Since the torsion element has only two degrees of freedom, Equations (B.42) to

(B.47) are once again utilized, i.e. ( ) ( )1 1 2 2x x xH x H xθ θ θ= + .

Recalling Equation (A.48) with G and Ip constants, the differential equation of motion

for a torsion element is given as:

( )2 2

2 2,

px xp x

mIGI m x t

x A t

θ θ∂ ∂− + =

∂ ∂ (B.59)


Equation (B.59) by a weighting function x

δθ and integrates along the element length:

( )2 2

2 2

0

, 0

l

px xp x x

mIGI m x t dx

x A t

θ θδθ

∂ ∂− + = =

∂ ∂ ∫ (B.60)

1 1ˆˆ , x

T θ2 2ˆ ,

xT θ

( ),x

p x t , , pG I m


364

The term 2

2

0

l

xp

GI dxx

θ∂

∂∫ in Equation (B.60) is now integrated by parts using Equation

(B.11) resulting in the following:

2

2 12 0

0 0

ˆ ˆl l

x x xp x x pl

GI dx T T GI dxx x x

θ δθ θδθ δθ

∂ ∂ ∂= + −

∂ ∂ ∂∫ ∫ (B.61)

where the torque T is defined in Equations (A.43). Substituting Equation (B.61) into

(B.60) and rearranging gives:

( )2

1 22 0

0 0

ˆ ˆ,

l lpx x x

p x x x x x l

mIGI dx dx m x t dx T T

x x A t

δθ θ θδθ δθ δθ δθ

∂ ∂ ∂+ = + +

∂ ∂ ∂ ∫ ∫ (B.62)

Comparing Equation (B.62) with (B.51), one can see a remarkable similarity between

both equations. The terms EA , m , u and P in Equation (B.51) are replaced by the

terms p

GI , pmI A ,

xθ and T in Equation (B.62), respectively; thus, one can

conclude that:

( )11 1 1 2 1 1 1 2 1 11

22 1 2 2 2 1 2 2 20 0 1 0 2

ˆ,

ˆ

l l lpx x

p x

x x

mIH H H H H H H H H TGI dx dx m x t dx

H H H H H H H H HA T

θ θ

θ θ

′ ′ ′ ′ + = + ′ ′ ′ ′

∫ ∫ ∫&&

&&

(B.63)

Integrating the left hand side of Equation (B.63), one gets the following matrix

equation governing the rotational motion about the longitudinal axis.

1 1

2 2

1 1 2 1

1 1 1 26

p px x

x x

GI mI l

l A

θ θ

θ θ

− + −

&&

&&( )1 1

20 2

ˆ,

ˆ

l

x

H Tm x t dx

H T

= + ∫ (B.64)


365

B.6 Structural-Beam Element

Combining Equation (B.19), (B.38) (B.57) and (B.64), one can write the stiffness and

mass matrix, as well as the force vector for a three-dimensional structural beam

element in local coordinates. The following matrix equation governing the complete

motion of the beam element, in beam element coordinates, results:

3 2 3 2

3 2 3 2

2

2

3 2

3 2

0 0 0 0 0 0 0 0 0 0

12 6 12 60 0 0 0 0 0 0

12 6 12 60 0 0 0 0 0

0 0 0 0 0 0 0

4 6 20 0 0 0 0

4 6 20 0 0 0

0 0 0 0 0

12 60 0 0

12 60 0

0 0

40

4.

z z z z

y y y y

p p

y y y

z z z

z z

y y

p

y

z

A A

l l

I I I I

l l l l

I I I I

l l l l

GI GI

El El

I I I

l l l

I I I

l l lE

A

l

I I

l l

I I

l l

GI

El

I

l

Isym

l

−

− − − − −

−

−

1

1

1

1

1

1

2

2

2

2

2

2

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

x

y

z

x

y

z

u

v

w

u

v

w

θ

θ

θ

θ

θ

θ


366

1

2 2

2 2

2

2

140 0 0 0 0 0 70 0 0 0 0 0

156 0 0 0 22 0 54 0 0 0 13

156 0 22 0 0 0 54 0 13 0

140 700 0 0 0 0 0 0

4 0 0 0 13 0 3 0

4 0 13 0 0 0 3

140 0 0 0 0 0420

156 0 0 0 22

156 0 22 0

1400 0

4 0

. 4

p p

p

u

l l

l l

I I

A A

l l l

l l lml

l

l

I

A

l

sym l

− −

− − −

+ −

&&

&1

1

1

1

1

2

2

2

2

2

2

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

x

y

z

x

y

z

v

w

u

v

w

θ

θ

θ

θ

θ

θ

=

&

&&

&&

&&

&&

&&

&&

&&

&&

&&

&&

( )( )( )( )( )( )( )( )( )( )( )( )

1

1

1

1

11

11

11

1 1

02 2

22

21

2

22

22

2

ˆ

, ˆ,

ˆ,

ˆ,

ˆ,

ˆ,

ˆ,

, ˆ

, ˆ

,ˆ

,ˆ

,ˆ

x

y

y

zz

x

yy

lz z

x

yy

zy

x

y

yz

z

PH p x t dx

QN p x t dx

QN p x t dx

TH m x t dx

MG p x t dx

G p x t dx M

H p x t dx P

N p x t dx QN p x t dx

QH m x t dx

TG p x t dx

MG p x t dx

M

− = +

−

∫ ext int

= + =

f f f (B.65)

Equation (B.65) can also be rewritten in finite element form as:

ku + mu = f&& (B.66)

where k and m are the local stiffness and mass matrix of the beam element, while u

and u&& are the local displacement and acceleration vectors, while f is the force vector.


367

B.7 Transformation from local to global axes

Until now, one has only ever concentrated on the local coordinate system of an

element; however, in order to assemble the mass and stiffness matrix for the entire

structure, we must first define each element in a global coordinate system. The local

right-hand coordinate system of an element is defined by x, y, and z, while the global

right-hand coordinate system for the entire structure is X , Y and Z as shown in

Figure B.9. Additionally, the six unit vectors parallel to the axes are also shown on the

diagram.

Figure B.9: Local and global coordinate system

In Figure B.10, the vector P can be represented in local or global coordinates as

follows:

x x y y z z x x y y z zp p p P P PP e e e E E E= + + = + + (B.67)

where, for example,x

p and x

e are the component and unit vector of P in the x

direction.

xe

ye yE

XxE

zE

ze

Yy

x

z

Z


368

Taking the dot product of Equation (B.67) with x

e one gets the following:

x x x x y y x z z x x x x y y x z z xp p p P P PP e e e e e e e E e Ε e E e⋅ = ⋅ + ⋅ + ⋅ = ⋅ + ⋅ + ⋅ (B.68)

This can also be written as follows:

x x x x y y x z z xp P P PE e E e E e= ⋅ + ⋅ + ⋅ (B.69a)

Similarly,

y x x y y y y z z yp P P PE e E e E e= ⋅ + ⋅ + ⋅ (B.69b)

z x x z y y z z z zp P P P= ⋅ + ⋅ + ⋅E e E e E e (B.69c)

Figure B.10: Vector P in local or global coordinates

Equation (B.69) can also be written in matrix form as:

xp y

p

yP

Xz

P

zp

Y

Z

x

z

y

xP

xey

e

yE

xE

zE

ze


369

0

x x x x y x z x x

y y x y y y z y y

z z x z y z z z z

p P P

p P P

p P P

⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅

e E e E e E

e E e E e E R

e E e E e E

(B.70a)

where 0R is known as the rotation matrix and is defined as:

0

x x x y x z

y x y y y z

z x z y z z

⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

e E e E e E

R e E e E e E

e E e E e E

(B.70b)

It should be noted that any quantity defined in terms of axes X , Y and Z can be

redefined in terms of the axes x, y, and z by pre-multiplying by the rotation matrix.

Initially, one takes a beam element whose x-axis lies along the X -direction with its y-

and z-axes lying along the Y and Z axes respectively. Then it moves by a series of

three rotations to come to its final position. The rotations are (1) a rotation α about the

y-axis; (2) a rotation β about the current z-axis; and (3) a rotation γ about the current

x-axis. Figure B.11 shows the rotation α of the member OA to OA′ about the z-axis.

The rotation matrix Rα of the member is obtained from Equation (B.70b) as:

( )

( )

o o

o o o

o o

cos cos90 cos 90

cos90 cos0 cos90

cos 90 cos90 cos

x x x y x z

y x y y y z

z x z y z z

α α α

α α αα

α α α

α α

α α

+ ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ −

e E e E e E

R e E e E e E

e E e E e E

(B.71)


370

Figure B.11: Rotation α of member OA about y-axis

Equation (B.71) is now solved using the following general expressions:

( )cos cos cos sin sinA B A B A B+ = − (B.72a)

( )cos cos cos sin sinA B A B A B− = + (B.72b)

resulting in the following:

cos 0 sin

0 1 0

sin 0 cos

α

α α

α α

− =

R (B.73)

hence,

x x

y y

z z

p P

p P

p P

R

α

αα

α

=

(B.74)

α

α

xe

α

yeαy

E

xE

zE

ze

α

x

, y Y

X

Z

z


371

Next, the member is rotated from OA′ to OA′′ by means of the rotation β about the z-

axis as shown in Figure B.12. It should be noted that the finished rotated position of

Figure B.11, now becomes the new datum position at which the new angles will be

measured. The rotation matrix Rβ is obtained from Equation (B.70b) as follows:

( )

( )

o o

o o

o o o

cos cos 90 cos90

cos 90 cos cos90

cos90 cos90 cos0

x x x y x z

y x y y y z

z x z y z z

β α β α β α

β α β α β αβ

β α β α β α

β β

β β

− ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ = + ⋅ ⋅ ⋅

e e e e e e

R e e e e e e

e e e e e e

(B.75)

Solving Equation (B.75) using (B.72), one gets

cos sin 0

sin cos 0

0 0 1

β

β β

β β

= −

R (B.76)

Figure B.12: Rotation β of member OA′ about z-axis

β

α

yeα

xE

zE

, z z

e eα β

x

Z

X

Yy

α

β

z

xe

α

xe

β

yE

yeβ


372

The effects of this second rotation are obtained by pre-multiplying the result of (B.74)

by Rβ, giving:

x x x

y y y

z z z

p p P

p p P

p p P

R R R

β α

β αβ β α

β α

= =

(B.77)

Finally, one rotates the element OA′′ about its own centroidal axis i.e. the x-axis, by

means of the rotation γ as illustrated in Figure B.13. As before, the finished rotated

position of Figure B.12 now becomes the new datum position at which the new angles

will be measured. The rotation matrix Rγ from Equation (B.70b) is given as:

( )

( )

o o o

o o

o o

cos0 cos90 cos90

cos90 cos cos 90

cos90 cos 90 cos

x x x y x z

y x y y y z

z x z y z z

β β β

β β βγ

β β β

γ γ

γ γ

⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ = − ⋅ ⋅ ⋅ +

e e e e e e

R e e e e e e

e e e e e e

(B.78)

Figure B.13: Rotation γ of member OA′′ about its centroidal axis (x-axis)

γ

β

α

x

Z

X

Yy

α

β

z

γ

yeα

xE

zE

ze

α

xe

α

yeγyE

yeβ

zeβ

zeγ

xe

βx

eγ


373

Solving Equation (B.78) using (B.72) gives

1 0 0

0 cos sin

0 sin cos

γ γ γ

γ γ

= −

R (B.79)

The effects of the third rotation are obtained by pre-multiplying the result of (B.77) by

Rγ, which gives:

0

x x x x

y y y y

z z z z

p p P P

p p P P

p p P P

R R R R R

γ β

γ βγ γ β α

γ β

= = =

where 0R is defined as:

0

1 0 0 cos sin 0 cos 0 sin

0 cos sin sin cos 0 0 1 0

0 sin cos 0 0 1 sin 0 cos

β β α α

γ γ β β

γ γ α α

− = − −

R

1 0 0 cos cos sin cos sin

0 cos sin sin cos cos sin sin

0 sin cos sin 0 cos

β α β β α

γ γ β α β β α

γ γ α α

− = − −

cos cos sin cos sin

cos sin cos sin sin cos cos cos sin sin sin cos

sin sin cos cos sin sin cos sin sin sin cos cos

β α β β α

γ β α γ α γ β γ β α γ α

γ β α γ α γ β γ β α γ α

− = − + + + − − +

(B.80)


374

B.8 Rotating the element equation from local to global coordinates

In this section, the author rotates the element equation from the local to global

coordinate system. Recalling Equation (B.66), the element equation in local

coordinates is defined as follows:

ku + mu = f&& (B.81)

where k and m are the local stiffness and mass matrix of the beam element, while u

and u&& are the nodal displacement and acceleration vectors and f is the force vector.

From Section B.6, it is found that the local displacement is transformed to a global

displacement as follows:

u TU= (B.82)

where the transformation matrix T is defined as:

0

0

0

0

0 0 0

0 0 0

0 0 0

0 0 0

R

RT

R

R

=

(B.83)

where each 0R represents a 3x3 matrix as given in Equation (B.80). Therefore,

Equation (B.81) can be rewritten in global coordinates as:

kTU + mTU = f&& (B.84)


375

Then, pre-multiplying Equation (B.84) by TT gives:

T T TT kTU + T mTU = T f&& (B.85a)

or

T =KU + MU = T f F&& (B.85b)

where K and M are the stiffness and mass matrix in global coordinates, respectively,

while F is force vector of the structure in global coordinates. Equation (B.85) is the

complete equation of motion of a beam element in the global coordinate system.

Equation (B.85a) can now be expanded using Equation (B.65) giving:

3 2 3 2

3 2 3 2

2

2

3 2

3 2

0 0 0 0 0 0 0 0 0 0

12 6 12 60 0 0 0 0 0 0

12 6 12 60 0 0 0 0 0

0 0 0 0 0 0 0

4 6 20 0 0 0 0

4 6 20 0 0 0

0 0 0 0 0

12 60 0 0

12 60 0

0 0

40

4.

z z z z

y y y y

p p

y y y

z z z

T

z z

y y

p

y

z

A A

l l

I I I I

l l l l

I I I I

l l l l

GI GI

El El

I I I

l l l

I I I

l l lE

A

l

I I

l l

I I

l l

GI

El

I

l

Isym

l

−

− − − − −

−

−

T

1

1

1

1

1

1

2

2

2

2

2

2

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

x

y

z

x

y

z

U

V

W

U

V

W

Θ

Θ Θ

Θ Θ Θ

T


376

2 2

2 2

2

2

140 0 0 0 0 0 70 0 0 0 0 0

156 0 0 0 22 0 54 0 0 0 13

156 0 22 0 0 0 54 0 13 0

140 700 0 0 0 0 0 0

4 0 0 0 13 0 3 0

4 0 13 0 0 0 3

140 0 0 0 0 0420

156 0 0 0 22

156 0 22 0

1400 0

4 0

. 4

p p

T

p

l l

l l

I I

A A

l l l

l l lml

l

l

I

A

l

sym l

− −

− − −

+ −

T T

&&1

1

1

1

1

1

2

2

2

2

2

2

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

x

y

z

x

y

z

U

V

W

U

V

W

Θ Θ Θ

= Θ Θ Θ

&&

&&

&&

&&

&&

&&

&&

&&

&&

&&

&&

( )( )( )( )( )( )( )( )( )( )( )( )

1

1

1

1

11

11

11

1 1

02 2

22

21

2

22

22

ˆ

, ˆ,

ˆ,

ˆ,

ˆ,

ˆ,

ˆ,

, ˆ

, ˆ,

ˆ,

ˆ,

ˆ

x

y

y

zz

x

yy

lz zT T

x

yy

zy

x

y

yz

z

PH p x t dx

QN p x t dx

QN p x t dx

TH p x t dx

MG p x t dx

G p x t dx M

H p x t dx P

N p x t dx QN p x t dx

QH p x t dx

TG p x t dx

MG p x t dx

M

− = +

−

∫T T ext int

2

= + =

F F F (B.86)

where U, V and W are the global displacements in the X, Y and Z-direction,

respectively, while ˆx

Θ , ˆyΘ and ˆ

zΘ are the global nodal rotations about the X, Y and

Z-axis, respectively. In addition, each dot over a symbol represents a derivative with

respect to time. The internal forces in Equation (B.86) are then eliminated by forcing

the joint into equilibrium during assembly, as shown in the next subsection.


377

B.9 Equilibrium of the joints - Assembly

The aim of this sub-section is to show that the internal forces between elements cancel

during assembly. Two rules must be satisfied to achieve the assembled global matrix.

Firstly, there must be compatibility of displacement at the element node i.e. all

members meeting at a joint must have the same displacement. Secondly, there must

be force equilibrium i.e. the sum of the internal forces meeting at a joint must balance

the external forces applied to that joint (Fish & Beltschko, 2007). As an example, the

author presents a beam of length L discretized into four beam elements of length l as

can be seen in Figure B.14a. An external force ( ),yp x t is then applied to the 2nd

beam element as shown in Figure B.14b.

1 3 4 52

Figure B.14: (a) Beam of length L discretized into four beam elements; (b) external

force applied to 2nd

element

Ignoring the mass of the beam, the global equation for the entire structure, denoted by

subscript s is constructed, with the aid of Equation (B.85b), as follows:

0s s s

− =K U F (B.87a)

L

l

element 1

( ),yp x t

(a)

(b)

element 2 element 3 element 4

1

2

1

2

1

2

1

2

,X U

,Y V


378

where

( )

1

Ee

s

e=

=∑K K (B.87b)

and the superscript in brackets in Equation (B.87b) refers to the element number.

Expanding Equation (B.87b) for our particular example give:

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )

1 1

11 121 1

1 1 2 2

21 22 11 12 2 2

2 2 3 33 321 22 11 12

3 3 4 44 4

21 22 11 12

4 4 5 521 22

0 0 0

0 0

00 0

0 0

0 0 0

+

− =+

+

K K U F

K K K K U F

U FK K K K

U FK K K K

U FK K

(B.88a)

or

1 111 12

2 221 22 23

3 331 33 34

4 443 44 45

5 554 55

0 0 0

0 0

00 0

0 0

0 0 0

− =

U FK K

U FK K K

U FK K K

U FK K K

U FK K

(B.88b)

As seen in Equation (B.88a), the global stiffness matrix of the entire structure is

formed by the addition of beam elements that share a common joint. The process

known as joint connectivity tells the user which elements share a common joint.

Next author applies equilibrium at the nodes in order to compute the internal forces in

the beam. The joint examined is node number 2. This node is connected to local node

2 of the 1st element and local node 1 of the 2

nd element as shown in Figure B.14a. In

accordance with Newton’s third law, the internal forces applied to the joint must be


379

equal and opposite to the internal forces applied at the ends of the beam. It is

important that the reader refers to Figures B.1, as this shows the sign convention for

internal forces acting on an element, while the sign conventions for internal forces

acting on a particular face are shown in Figure B.2. A free-body diagram of node 2 is

shown in Figure B.15

Figure B.15: Free-body diagram of node number 2 and its internal forces

Summing the internal and external forces acting on node 2 can be written as follows:

( ) ( ) ( ) ( )10


0 , 0l

y y yQ l Q N x p x t dx− − =∫

or ( ) ( )2 1 10

ˆ ˆ , 0l

y y yQ Q N x p x t dx− − =∫ (B.89a)

( ) ( ) ( ) ( )10


0 , 0l

z z yM l M G x p x t dx− − =∫

or ( ) ( )2 1 10

ˆ ˆ , 0l

z z yM M G x p x t dx− − =∫ (B.89b)

From inspection of Equation (B.89), one can see that if there are no external forces

acting on a particular node, then the internal forces between elements must be equal:

2 1ˆ ˆ 0y yQ Q− = then 2 1

ˆ ˆy yQ Q= (B.90a)

2 1ˆ ˆ 0

z zM M− = then 2 1

ˆ ˆz z

M M= (B.90a)

Node 2

2ˆ

zM

1ˆ

zM

2ˆ

yQ1

ˆy

Q

2ˆ

zM1

ˆz

M

1ˆ

yQ 2

ˆy

Q1i = 2i =Elem 1 Elem 22i = 3i =

( ),yp x t

2ˆ

yQ1

ˆyQ

1ˆ

zM2

ˆz

M

2i =


380

B.10 Applying rotations to axial elements

This section rotates an axial element that is defined in Section B.4 from local to global

coordinates, which is used in the development of the wheel-rail contact elements in

Chapter 4. Additionally, the author omits the mass matrix from the following

equations as this axial element i.e. wheel-rail contact element has a zero mass;

therefore, Equation (B.85) becomes:

T TT kTU = T f (B.91a)

or

KU = F (B.91b)

where K is the global stiffness matrix, U is the global displacement vector, f is the

local force vector and T is the transformation matrix. By selecting the first three terms

on the left-hand side of Equation (B.91a), one can rewrite the global stiffness matrix

of the axial element are follows:

011 120

00

021 220

00

0 0 00 00 0 0

0 0 00 0 0 00 0 0

0 0 00 00 0 0

0 0 00 0 0 00 0 0

T

L L

T

T

T

L L

T

Rk kR

RRT kT =

Rk kR

RR

00 11 0 12

0

00 21 0 22

0

0 0 00 0

0 0 00 0 0 0

0 0 00 0

0 0 00 0 0 0

T T

L L

T T

L L

=

RR k R k

R

RR k R k

R

11 120 11 0 0 12 0

21 220 21 0 0 22 0

0 00 0

0 0 0 00 0 0 0

0 00 0

0 0 0 00 0 0 0

T TL LL L

T TL LL L

= =

K KR k R R k R

K KR k R R k R (B.92)


381

where 0R is defined in Equation (B.80), Lij

K (i, j =1, 2) denotes the axial element

(3 3× ) stiffness matrix in the global coordinate system and Lijk (i, j =1, 2) denotes the

axial element (3 3× ) stiffness matrix in the local coordinate system, which can be

written from Equation (B.65) as:

11 12 21 22

0 0

0 0 0

0 0 0

L L L L

EA l = − = − = =

k k k k (B.93)

Since it is assumed that the axial element never undergoes a rotation γ about the x-

axis, it is given a value of zero; therefore, Equation (B.80) becomes:

0

cos cos sin cos sin

sin cos cos sin sin

sin 0 cos

R

β α β β α

β α β β α

α α

− = −

(B.94)

Substituting Equation (B.94) into the first quadrant of Equation (B.92) gives:

11 0

1 0 0 C C S C S C C S C S

0 0 0 S C C S S 0 0 0

0 0 0 S 0 C 0 0 0

L

EA EA

l lk R

β α β β α β α β β α

β α β β α

α α

− − = − =

(B.95)

where, for convenience, C cosα α= , S sinα α= , C cosβ β= and S sinβ β= . Now,

combining Equation (B.95) with the transpose of Equation (B.94), one gets:

0 11 0 11

C C S C S C C S C S

S C 0 0 0 0

C S S S C 0 0 0

T

L L

EA

lR k R K

β α β α α β α β β α

β β

β α β α α

− − = = −


382

2 2 2

2

2 2 2

C C C S C C C S

C S C S C S S

C C S C S S C S

EA

l

β α β β α β α α

β β α β β β α

β α α β β α β α

−

= − − −

(B.96)

For a vertical spring element as shown in Figure B.16a, one assumes that the rotation

β about the z-axis is equal to o90 , while the rotation α about the y-axis is equal to

zero; therefore, Equation (B.96) becomes:

0 11 0 11

0 0 0

0 1 0

0 0 0

T

L L

EA

lR k R K

= =

(B.97)

For a longitudinal spring element presented in Figure B.16b, one assumes that the

rotation β about the z-axis is equal to o0 while the rotation α about the y-axis remains

equal to zero; therefore, Equation (B.96) becomes:

0 11 0 11

1 0 0

0 0 0

0 0 0

T

L L

EA

lR k R K

= =

(B.98)

Finally, for the lateral spring element as can be seen in Figure B.16c, one assumes that

the rotation β about the z-axis is equal to o0 , while the rotation α about the y-axis is

equal to o270 so that the lateral spring is aligned with the positive z-axis; therefore,

Equation (B.96) becomes:

0 11 0 11

0 0 0

0 0 0

0 0 1

T

L L

EA

lR k R K

= =

(B.99)


383

Figure B.16: Spring element aligned with the global Y, X and Z-axis

(b) X

Y

Z

x

21

y

z

(c) X

Y

Z x

2

1

y

z

(a) X

Y

Z

x

2

1

y

z


384

B.11 Equation for the axial extension of a spring element

In this section, the author develops an equation for the axial extension of a spring

element, which is used in Chapter 4 to calculate the contact force between the wheel

and the rail. Recalling Equation (A.8), the strain along the element length is equal to:

xx

u

xε

∂=

∂ (B.100a)

or

change in length of the element

original length of the elementxx

l

lε

∆= = (B.100b)

Substituting Equation (B.46a) into (B.100a) then gives:

1 21xx

x xu u

x l x lε

∂ ∂ = − +

∂ ∂

1 2

1 1u u

l l

= − +

(B.101)

Next, substituting Equation (B.100b) into (B.101), one can relate the change in length

of the spring element i.e. the extension as follows:

1 2

1 1lu u

l l l

∆ = − +

1 2 extensionl u u∆ = − + = (B.102)

To transform the local displacements in Equation (B.102) to global displacements, the

author must use Equation (B.82), which is rewritten here as:

u TU= (B.103)


385

where the transformation matrix T is defined again as:

0

0

0

0

0 0 0

0 0 0

0 0 0

0 0 0

R

RT

R

R

=

(B.104)

In Section B.10, one found that since the spring element is only rotated about two

axes, firstly about the y-axis and secondly about the z-axis, R0 can be rewritten with

the aid of Equation (B.94) as follows:

0

cos cos sin cos sin

sin cos cos sin sin

sin 0 cos

R

β α β β α

β α β β α

α α

− = −

(B.105)

Substituting Equation (B.105) into (B.104) and then into Equation (B.103) gives the

local displacements in the global coordinate system as follows:

0

0

0

0

0 0 0

0 0 0

0 0 0

0 0 0

=

R

Ru U

R

R

(B.106a)

1 1

1 1

1 1

2 2

2 2

2 2

cos cos sin cos sin 0 0 0

sin cos cos sin sin 0 0 0

sin 0 cos 0 0 0


0 0 0 sin cos cos sin sin

0 0 0 sin 0 cos

u U

v V

w W

u U

v V

w W

β α β β α

β α β β α

α α

β α β β α

β α β β α

α α

− −

= −

−

(B.106b)


386

Expanding only the 1st and 4

th row of Equation (B.106b), the displacement of the

spring element in the local x direction can be written as follows:

1

1

1 1

2 2

2

2



U

V

u W

u U

V

W

β α β β α

β α β β α

−

= −

(B.107a)

or with a change of notation as:

1

1

1 1

2 2

2

2



Lx

Ly

Lz

Lx

Ly

Lz

U

U

u U

u U

U

U

β α β β α

β α β β α

−

= −

(B.108)

where Lxi

U (i = 1, 2) is the global horizontal displacement, Lyi

U (i = 1, 2) is the

global vertical displacement and Lzi

U (i = 1, 2) is the global lateral displacement of

the spring element. With the aid of Equation (B.102) and (B.108), the extension on

the spring element in the global coordinate system can be written as:

[ ]1

1

1

extension cos cos sin cos sin

Lx

Ly

Lz

U

U

U

β α β β α

= − −

[ ]2

2

2

cos cos sin cos sin

Lx

Ly

Lz

U

U

U

β α β β α

+ −

(B.109)

As a test, let α be equal to 0o and β be equal to 90

o, then Equation (B.109) reduces to:

[ ]1

1

1

extension 0 1 0

Lx

Ly

Lz

U

U

U

= −

[ ]2

2

2

0 1 0

Lx

Ly

Lz

U

U

U

+

(B.110)

Appendix C – Natural Frequencies and Modal Shapes for a beam

387

Appendix C

Natural Frequencies and Modal Shapes for a beam

C.1 Mode shape for any beam

In this Appendix, the author calculates the natural frequencies and modal shapes of

single-span beams as shown in Figure C.1 under different end restraints.

Figure C.1: Single span beam and coordinate system

From Equation (A.22) with z

EI constant and omitting the damping term, the

differential equation governing the free vibration of a beam is expressed as:

4 2

4 20

v vEI m

x t

∂ ∂+ =

∂ ∂ (C.1)

z

x

y, v

0 m, EIz

L


388

where E is Young’s modulus of elasticity, I is the second moment of area and m is the

mass per unit length of the beam. It should also be noted that beam deflection ( ),v x t

is expressed as a function of both time t and position x along the span.

According to Biggs (1964), since the right-hand side of Equation (C.1) is set equal to

zero, one can define the beam deflection of the n-th mode by:

( ) ( ) ( ),n n nv x t f t xφ= (C.2)

where ( )nf t is a time function, ( )n xφ is the characteristic shape. From Equation

(C.2), one can also write the following relationships:

( ) ( )( )

22

2 2

,n nn

d f tv x tx

t dtφ

∂=

∂ and

( )( )

( )4 4

4 4

,n n

n

v x t d xf t

x dx

φ∂=

∂ (C.3)

Substituting Equation (C.3) into (C.1) gives

( )( ) ( )

( )4 2

4 20

n n

n n

d x d f tEIf t m x

dx dt

φφ+ = (C.4a)

or

( )( )

( )( )4 2

4 2

1n n

n n

d x d f tEI

m x dx f t dt

φ

φ= − (C.4b)

From inspection of Equation (C.4b), one can conclude that the left side varies only

with x, while the right side varies only with t; hence, each expression can be set equal

to a constant term, H. We can now rewrite the left-hand side and the right-hand side

of Equation (C.4b) separately as follows:


389

( ) ( )2

20

n n

df t Hf t

dt+ = (C.5a)

( ) ( )4

40

n n

d mHx x

dx EIφ φ− = (C.5b)

If H is zero or negative, the solution, ( )nf t , of Equation (C.5a) is not periodic in time

and so we would not be solving the natural period problem, which is associated with

the modal shape. Therefore H must be positive and the solution to Equation (C.5a) is

as follows:

( ) 1 2sin cosnf t C Ht C H t= + (C.6)

Clearly, H is the frequency, n

ω . We can now rewrite Equation (C.5b) as:

( ) ( )24

40n

n n

mdx x

dx EI

ωφ φ− = (C.7)

Let the solution to Equation (C.7) be

( ) px

n x Ceφ = (C.8)

where C is constant. Substituting Equation (C.8) into (C.7) gives

24 0px pxn

mp Ce Ce

EI

ω− = or

24 0n

mp

EI

ω− = (C.9)


390

Factorizing Equation (C.9), one gets:

( )( )2 2 2 2 0n n

p m EI p m EIω ω− + = (C.10a)

or

( )( )( )( )2 2 2 24 4 4 4 0n n n n

p m EI p m EI p i m EI p i m EIω ω ω ω− + − + = (C.10b)

where the following mathematical expression are used to evaluate the roots

( )( )2 2p a p a p a− = − + (C.11a)

( ) ( )2 2p a p ia p ia+ = + − (C.11b)

1i = − (C.11c)

Equation (C.10b) is substituted into (C.8) and thus gives rise to the following solution

for ( )n xφ

( ) sin cos sinh coshn n n n n n n n n

x a x a x a x a xφ = + + +A B C D (C.12a)

where

24n na m EIω= (C.12b)

, A B, C, D are constant values evaluated by using the boundary conditions of the beam.

Equation (C12a) is valid for any type of end restraint. The 1st, 2

nd and 3

rd derivatives

of Equation (C.12a) are as follows:

[ ]( ) cos sin cosh sinhn n n n n n n n n n

dx a a x a x a x a x

dxφ = − + +A B C D (C.12c)

[ ]2

2

2( ) sin cos sinh cosh

n n n n n n n n n n


dxφ = − − + +A B C D (C.12d)

[ ]3

3

3( ) cos sin cosh sinh

n n n n n n n n n n


dxφ = − + + +A B C D (C.12e)


391

C.2 Cantilever Beam –Natural Frequencies and Mode Shapes

In the following example, the author examines the free vibration of a cantilever beam

that is supported at the origin; hence, the boundary conditions are as follows:

0v

vx

∂= =

∂ at x = 0 (C.13a)

2 3

2 30

v vEI EI

x x

∂ ∂= − =

∂ ∂ at x = L (C.13b)

where v is the deflection of the beam, v x∂ ∂ is the slope as shown in Equation (A.2),

2 2EI v x∂ ∂ is the bending moment as shown in Equation (A.16), and 3 3EI v x− ∂ ∂ is

the shear force as shown in Equation (A.19). The function ( )n

xφ satisfies the same

boundary condition as ( , )v x t . This can be seen by substituting Equation (C.2) into

Equation (C.13). Evaluating ( )n

xφ at x = 0 using Equation (C.12a) and the first

derivative of that equation gives:

( )0 0n n nφ = = +B D n n

= −D B

(0) 0n n n n n

da a

dxφ = = +A C

n n= −C A (C.14)

Substituting Equation (C.14) back into (C.12a) gives the following:

( ) ( )( ) sin sinh cos coshn n n n n n nx a x a x a x a xφ = − + −A B (C.15)

Similarly, one substitutes x = L into the second and third derivative of ( )n

xφ using

Equation (C.12):


392

( ) ( )2

2 2

2( ) sin sinh cos cosh 0

n n n n n n n n n

dL a a L a L a a L a L

dxφ = − − − − =A +B (C.16a)

( ) ( )3

3 3

3( ) cos cosh sin sinh 0

n n n n n n n n n


dxφ = − − + − =A B (C.16b)

Since ,n nA B cannot both be equal to zero for a non-trivial value of

nφ , then the

determinant of the coefficients must be zero. The result is:

sin sinh cos cosh0

cos cosh sin sinh

n n n n

n n n n

a L a L a L a L

a L a L a L a L

− − − −=

− − − (C.17)

The formula for evaluating the determinants of a 2x2 matrix is:

a bad bc

c d= − (C.18a)

Equation (C.17) therefore becomes:

( )( ) ( )( )sin sinh sin sinh cos cosh cos cosh 0n n n n n n n n

a L a L a L a L a L a L a L a L− − − − − − − − = (C.18b)

which can also be written as follows:

( ) ( )sin sin sinh sinh sin sinhn n n n n na L a L a L a L a L a L− − − −

( ) ( )cos cos cosh cosh cos cosh 0n n n n n na L a L a L a L a L a L+ − − + − − = (C.18c)

Tidying up Equation (C.18c), one gets:

2 2sin sin sinh sin sinh sinhn n n n n n

a L a L a L a L a L a L− + − +

2 2cos cos cosh cos cosh cosh 0n n n n n n

a L a L a L a L a L a L− − − − = (C.18d)


393

Rearranging Equation (C.18d) gives:

( ) ( )2 2 2 22cos cosh sin cos cosh sinh 0n n n n n n

a L a L a L a L a L a L+ + + − = (C.18e)

From trigonometry one has the following:

2 2sin cos 1A A+ = (C.18f)

2 2cosh sinh 1A A− = (C.18g)

Substituting Equation (C.18f) and (C.18g) into Equation (C.18e) gives:

2cos cosh 2 0n n

a L a L + =

which reduces to

cos cosh 1 0n n

a L a L + = (C.18h)

Solving Equation (C.16a) for nA gives:

( ) ( )2 2sin sinh cos cosh 0n n n n n n n na a L a L a a L a L− − − − =A +B (C.19a)

( ) ( )2 2sin sinh cos coshn n n n n n n na a L a L a a L a L− − = − − −A B (C.19b)

( )( )

cos cosh cos cosh

sin sinh sin sinh

n n n nn n n

n n n n

a L a L a L a L

a L a L a L a L

− − += − = −

− − + A B B (C.19c)

Substituting Equation (C.19c) into Equation (C.15) then gives

( ) ( )cos cosh

( ) sin sinh cos coshsin sinh

n n

n n n n n n n

n n

a L a Lx a x a x a x a x

a L a Lφ

+= − − + −

+ B B (C.20)


394

The undetermined amplitude nB is then evaluated by normalizing ( )n xφ as follows:

( )2

0

L

nx dx Lφ =∫ (C.21)

As an exercise, nB is evaluated for the cantilever beam in Section C.6. It should be

noted that Equation (C.21) can have a value of plus or minus nB ; however, in this

thesis one uses the negative value. In addition, the modes are orthogonal:

( ) ( )0

L

n i inx x dx Lφ φ δ=∫ (C.22)

The natural frequencies,n

ω , of the cantilever beam are related to the roots n

a of

Equation (C.18h) through Equation (C.12b). In order to compute the roots n

a of

Equation (C.18h), this equation is rewritten as follows:

cos cosh 1 0n n

β β + = (C.23a)

n na Lβ = (C.23b)

The roots n

β of Equation (C.23a) are got by plotting the following function:

( ) cos cosh 1β β βΨ = + (C.24)

in Figure C.2. The roots occur when Ψ crosses the β-axis. These roots are

approximated by the points where cos β crosses the β-axis as can be seen in Figure

C.2. The roots of cos β are giving by the following equation, as stated in Biggs

(1964):

2 1, 1, 2, 3, ...

2n

nnβ π

− = =

(C.25)


395

-20

-15

-10

-5

0

5

10

0 2 4 6 8 10 12

ββββ

True

Cos

Figure C.2: Comparing the roots of Ψ and cos β graphically

In Table C1, the author presents βn, n = 1, 6 evaluated both exactly and approximately

for the cantilever beam. For n = 1, the approximate value is unsuitable as the error is

over 16%. For n > 1, the approximate roots are suitable because the differences

between the evaluated and approximate roots are less than 0.5%.

Table C1: First 6 roots of a cantilever beam

n Ψ (β) = 0 cos 0β = % Diff

1 1.875190 1.570796 16.23270%

2 4.694091 4.712389 0.38981%

3 7.854757 7.853982 0.00987%

4 10.995541 10.995574 0.00030%

5 14.137168 14.137167 0.00001%

6 17.278760 17.278760 0.00000%

Using Equation (C.20), a plot of the first 6 mode shapes of the cantilever beam can be

seen in Figure C.3. Substituting Equation (C.23b) into (C.12b) and rearranging in

3

2

π 5

2

π 7

2

π

2

π

Ψ (Ψ (Ψ (Ψ (ββββ)))) cos ββββ


396

terms of the natural frequencies n

ω , one gets the following relationship (measured in

rad/sec):

2

2

nn

EI

L m

βω = (C.26)

Using Table C1, one can write the first natural frequency 1ω using Equation (C.26) as

follows:

( )( )2

1 2

1.87519 EI

L m

π πω = (C.27a)

Alternatively, substituting Equation (C.25) into (C.26) gives approximate natural

frequencies for the cantilever beam, which are most suitable when n > 1.

( )2 2

2

1 2n

n EI

L m

πω

−= (C.27b)

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Dimensionless length

n = 1 n = 2 n = 3 n = 4 n = 5 n = 6

Figure C.3: First 6 mode shapes of the cantilever beam


397

C.3 Fixed-Fixed Beam – Natural Frequencies and Mode Shapes

In this example, the free vibration of a fixed-fixed beam is examined using a similar

methodology to the previous section; thus the boundary conditions are as follows:

0v

vx

∂= =

∂ at x = 0 (C.28a)

0v

vx

∂= =

∂ at x = L (C.28b)

where v is the deflection of the beam, v x∂ ∂ is the slope. The function ( )n

xφ satisfies

the same boundary conditions as ( , )v x t , namely Equation (C.28). Substituting x = 0

into Equation (C.12a) and into the first derivative of that Equation (C.12c), gives

( )0 0n n nφ = = +B D n n

= −D B

(0) 0n n n n n

da a

dxφ = = +A C

n n= −C A (C.29)

Substituting Equation (C.29) back into (C.12a) gives the following:

( ) ( )( ) sin sinh cos coshn n n n n n nx a x a x a x a xφ = − + −A B (C.30)

Equally, one substitutes x = L into Equation (C.30) and into the first derivative of that

equation giving:

( ) ( )( ) sin sinh cos cosh 0n n n n n n nL a L a L a L a Lφ = − + − =A B (C.31a)

( ) ( )( ) cos cosh sin sinh 0n n n n n n n n n


dxφ = − + − − =A B (C.31b)


398

Since ,n nA B cannot both be equal to zero for a non-trivial value of

nφ , then the

determinant of the coefficients must be zero, resulting in

sin sinh cos cosh0

cos cosh sin sinh

n n n n

n n n n

a L a L a L a L

a L a L a L a L

− −=

− − − (C.32)

which reduces to

cos cosh 1 0n n

a L a L − = (C.33)

Solving Equation (C.31a) for nA , one gets:

( ) ( )sin sinh cos cosh 0n n n n n na L a L a L a L− + − =A B (C.34a)

( ) ( )sin sinh cos coshn n n n n na L a L a L a L− = − −A B (C.34a)

( )( )cos cosh cos cosh

sin sinh sin sinh

n n n nn n n

n n n n

a L a L a L a L

a L a L a L a L

− − += − =

− + A B B (C.34c)

Substituting Equation (C.34c) into Equation (C.30) then gives

( ) ( )cos cosh


n n

n n n n n n n

n n


a L a Lφ

− += − + −

+ B B (C.35)

The undetermined amplitude nB is evaluated by normalizing ( )n xφ from Equation

(C.21); thus nB is given a value of unity. The evaluation of the natural

frequencies,n

ω , of the fixed-fixed beam are related to the roots n

a of Equation (C.33)

through Equation (C.12b); thus one computes the roots n

a of Equation (C.33) as

follows:


399

cos cosh 1 0n n

β β − = (C.36a)

where

n na Lβ = (C.36b)

The roots n

β of Equation (C36a) are got by plotting the following function

( ) cos cosh 1β β βΨ = − (C.37)

in Figure C.4. The roots occur when Ψ crosses the β-axis. These roots are

approximated by the points where cos β crosses the β-axis as can be seen in Figure

C.4. The roots of cos β are giving by the following equation:

2 1, 1, 2, 3, ...

2n

nnβ π

+ = =

(C.38)

-10

-5

0

5

10

0 2 4 6 8 10 12

ββββ

True

Cos

Figure C.4: Comparing the roots of Ψ and cos β graphically

Table C2 shows βn, n = 1, 6 evaluated both exactly and approximately for the fixed-

fixed beam. It can be clearly seen from Table C2 that the approximate roots have a

3

2

π 5

2

π 7

2

π

2

π

Ψ (Ψ (Ψ (Ψ (ββββ))))

cos ββββ


400

striking similarity to the exact roots. A plot of the first 6 mode shapes of the fixed-

fixed beam is presented in Figure C.5.

Table C2: First 6 roots of a fixed-fixed beam

n Eq. (37) Eq. (38) % Diff

1 4.730041 4.712389 0.37319%

2 7.853205 7.853982 0.00989%

3 10.995610 10.995574 0.00032%

4 14.137170 14.137167 0.00002%

5 17.278760 17.278760 0.00000%

6 20.420352 20.420352 0.00000%

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0


n = 1 n = 2 n = 3 n = 4 n = 5 n = 6

Figure C.5: First 6 mode shapes of the fixed-fixed beam

Substituting Equation (C.38) into (C.26) gives approximate natural frequencies for the

fixed-fixed beam as follows:

( )2 2

2

1 2n

n EI

L m

πω

+= n = 1, 2, 3 … (C.39)


401

C.4 Simply Supported Beam – Natural Frequencies and Mode Shapes

As a final example, the free vibration of a simply supported beam is examined. The

boundary conditions are as follows:

2

20

vv EI

x

∂= =

∂ at x = 0 (C.40a)

2

20

vv EI

x

∂= =

∂ at x = L (C.40b)

where v is the deflection of the beam, 2 2EI v x∂ ∂ is the bending moment. The

function ( )n

xφ satisfies the same boundary conditions as ( , )v x t , namely Equation

(C.40). Substituting Equation (C.12a) into Equation (C.40a) and into (C.40b) gives

( )0 0n n nφ = = +B D n n

= −D B

22 2

2(0) 0

n n n n n

da a

dxφ = = − +B D

n n=D B (C.41)

From Equation (C.41), one can conclude that both nB and

nD must be equal zero;

thus, Equation (C.12a) can be rewritten as follows:

( ) sin sinhn n n n n

x a x a xφ = +A C (C.42)

Substituting x = L into Equation (C.42) and its second derivative, one gets

( ) sin sinh 0n n n n n

L a L a Lφ = + =A C (C.43a)


402

22 2

2( ) sin sinh 0

n n n n n n

dL a a L a L

dxφ = − + =A C (C.43b)

Adding and subtracting Equation (C.43a) to (C.43b) after first cancelling the 2

na term

in Equation (C.43b) gives the following relationship:

2 sin 0n n

a L =A

2 sinh 0n n

a L =C (C.44b)

Since sinhn

a L cannot be equal to zero, then 0n

=C . Furthermore 0n

≠A , otherwise

the beam would experience no vibration; thus the roots are as follows:

sin 0β = (C.45a)

where

, 1, 2, 3, ...n n

a L n nβ π= = = (C.45b)

Since 0n n n

= = =B C D , then with the aid of Equation (C.45b), one can rewrite

Equation (C.12a) as follows, where nA is arbitrary.

( ) sinn n

n xx

L

πφ = A (C.46)

However in order to satisfy Equation (C.21), one must modify Equation (C.46) as

follows (as an exercise nA is evaluated in Section C.6):


403

( ) 2 sinn

n xx

L

πφ = (C.47a)

where

2n

=A (C.47b)

Figure C.6 plots the first 6 mode shapes of the simply supported beam.

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0


n = 1 n = 2 n = 3 n = 4 n = 5 n = 6

Figure C.5: First 6 mode shapes of the simply supported beam

Finally, substituting Equation (C.45b) into (C.26) gives the natural frequencies for the

simply supported beam as follows:

2 2

2n

n EI

L m

πω = (C.48)


404

C.5 Dimensionless speed ratio αααα for a simply supported beam

In this section, the dimensionless speed ratio α is defined. This parameter is used

throughout this thesis as a means of measuring the current operating vehicle speed

against a critical operating speed. The ratio of first natural frequency of the vehicle

1Ω to the first natural frequency of the bridge 1ω is called α and is written as:

1

1

αω

Ω= (C.49)

where

1

c

L

πΩ = (C.50)

The natural frequency of the bridge 1ω in Equation (C.49) is then replaced by 12 fπ ,

where 1f is in cycles per second giving:

1

12 f

απ

Ω= (C.51)

Substituting Equation (C.50) into (C.51) gives the following equation:

1

1 1 12 2 2 cr

c c c

f f L f L c

πα

π π

Ω= = = = (C.52)

where

12crc f L= (C.53)

Equation (C.53) relates the current operating speed c of the vehicle against its critical

operating speed crc traversing a simply supported beam.


405

C.6 Undetermined amplitude evaluated by normalization

As stated in previous sub-sections, the undetermined amplitude is evaluated by

normalizing the displacement function ( )n xφ as follows:

( )2

0

L

nx dx Lφ = ∫ (C.54)

One first begins with a simple model i.e. the normalization of the simply supported

beam. Recalling Equation (C.46) as follows:

( ) sinn n

n xx

L

πφ = A (C.55)

Squaring Equation (C.55) then gives:

[ ]2

2 2 2( ) sin sinn n n

n x n xx

L L

π πφ

= = A A (C.56)

Integrating Equation (C.56), with limits from 0 to L, one gets:

[ ]2 2 2

0 0

( ) sin

L L

n n

n xx dx dx

L

πφ

=

∫ ∫A (C.57a)

( )2

0

sin 2

2 4

L

n

L n x Lx

n

π

π

= −

A (C.57 b)

( )2 2sin 2

2 4 2n n

L n L LL LL

n

π

π

= − = =

A A (C.57c)


406

From inspection of Equation (C.57c), the reader can see that nA must be equal to

2± in order to satisfy Equation (C.54). In this thesis, the author uses 2+ . Using

real numbers, one will now show that Equation (C.57c) is true for a simply supported

beam by assuming the beam has a length L of 7.62 m and n has a value of 1; thus,

giving:

[ ]( )2 2

0

sin 2( )

2 4

L

n n

L n L LLx dx

n

πφ

π

= −

∫ A

( )2

7.62*27.62sin

7.62 7.622 7.62

2 4

π

π

= − =

(C.58)

In the case for a simply supported beam, nA is fully analytically. However, for the

cantilever beam, due to the complicated formula for the mode, the author will

integrate ( )n

xφ partially analytically and partially numerically. In order to evaluate

the undetermined amplitude nB for the cantilever beam, one normalises its

displacement function. Recalling Equation (C.20) as follows:

( ) ( )cos cosh


n nn n n n n n n

n n


a L a Lφ

+= − − + −

+ B B (C.59a)

[ ]sin sinh cos coshn n n n n n n

C a x C a x a x a x= − + −B (C.59b)

whereby

cos cosh

sin sinh

n nn

n n

a L a LC

a L a L

+= −

+ (C.59c)


407

Squaring Equation (C.59b) then gives

2 2 2 2 2( ) sin sin sinh sin cos sin coshn n n n n n n n n n n n n

x C a x C a x a x C a x a x C a x a xφ = − + −B

2 2 2sinh sin sinh sinh cos sinh coshn n n n n n n n n n n

C a x a x C a x C a x a x C a a x− + − +

2cos sin cos sinh cos cos coshn n n n n n n n n

C a x a x C a x a x a x a x a x+ − + −

2cosh sin cosh sinh cosh cos coshn n n n n n n n n

C a x a x C a x a x a x a x a x− + − + (C.60)

Simplifying Equation (C.60) gives

2 2 2 2 2 2 2( ) sinh sin 2 sinh sinn n n n n n n n n

x C a x C a x C a x a xφ = −B +

2 cos sin 2 cosh sin 2 cosh sinhn n n n n n n n n

C a x a x C a x a x C a x a x+ − +

2 22 sinh cos 2cosh cos cos coshn n n n n n n

C a a x a x a x a x a x− − + + (C.61)

Next, one integrates Equation (C.61) with limits from 0 to L; thus giving:

2 2 2 2 2 2 2

0 0 0 0

( ) sinh sin 2 sinh sin

L L L L

n n n n n n n n nx dx C a xdx C a xdx C a x a xφ

= + −

∫ ∫ ∫ ∫B

0 0 0

2 cos sin 2 cosh sin 2 cosh sinh

L L L

n n n n n n n n nC a x a xdx C a x a xdx C a x a xdx+ − +∫ ∫ ∫

2 2

0 0 0 0

2 sinh cos 2 cosh cos cos cosh

L L L L

n n n n n n nC a a xdx a x a xdx a xdx a xdx

− − + +

∫ ∫ ∫ ∫

2 2 2 22

0 0

sinh 2 sin 2

2 4 2 4

L L

n n n n n nn

n n

C x C a x C x C a x

a a

= − + + −

B

2 2

0

cos sinh cosh sinL

n n n n n n

n n

C a x a x C a x a x

a a

+ −


408

2 2

0 0 0

cos cos cosh sin sinh coshL L L

n n n n n n n n n n

n n n n

C a x C a x a x C a x a x C a x

a a a a

+ − + − +

0 0

sin sinh cos cosh cosh sin cos sinhL L

n n n n n n n n n n

n n n n

C a x a x C a x a x a x a x a x a x

a a a a

+ − − + − −

0 0

sin 2 sinh 2

2 4 2 4

L L

n n

n n

a x a xx x

a a

+ + + +

(C.62)

Expanding Equation (C.25) and simplifying gives:

2 22 2

0

sinh 2 sin 2( )

4 4

L

n n n nn n

n n

C a L C a Lx dx L

a aφ

= + −

∫ B

2 2cos sinh cosh sinn n n n n n

n n

C a L a L C a L a L

a a+ −

2 2cos coshn n n n

n n

C a L C a L

a a− +

2 sin sinhn n

n

C a L a L

a−

cosh sin cos sinhn n n n

n n

a L a L a L a L

a a− −

sin 2 sinh 2

4 4

n n

n n

a L a L

a a

+ +

(C.63a)

or

2 2 2

2 2

0

cosh sinh cos sin cos sinh( )

2 2

L

n n n n n n n n nn n

n n n

C a L a L C a L a L C a L a Lx dx L

a a aφ

= + − +

∫ B

2 2 2cosh sin cos coshn n n n n n n

n n n

C a L a L C a L C a L

a a a− − +

2 sin sinh cosh sin cos sinhn n n n n n n

n n n

C a L a L a L a L a L a L

a a a− − −

cos sin cosh sinh

2 2

n n n n

n n

a L a L a L a L

a a

+ +

(C.63b)

Recalling the normalization equation in Equation (C.54), one must now find nB such

that the right hand side of Equation (C.63b) is equal to L.


409

With the aid of real numbers, the author will show that the right hand side of Equation

(C.63b) is equal to the length of the beam. Using the exact same beam properties as

Section 5.3.1, the beam has a length L of 7.62, a mass per unit length m of 46 kg/m,

and a bending stiffness EI of 6 29.47 10 Nm .× Then by using Equations (C.27a),

(C.12b) and (C.59c), the first natural frequency of the beam 1ω is equal to 27.477

rad/sec, 1a has a value of 0.24609 and C1 has a value of -0.73410. Inputting these

constants into Equation (C.63b) for n = 1 then gives:

2 2 22 2 1 1 1 1 1 1 1 1 1

1 1

1 1 10


2 2

LC a L a L C a L a L C a L a L

x dx La a a

φ

= + − +

∫ B

2 2 2

1 1 1 1 1 1 1

1 1 1

cosh sin cos coshC a L a L C a L C a L

a a a− − +

1 1 1 1 1 1 1

1 1 1

2 sin sinh cosh sin cos sinhC a L a L a L a L a L a L

a a a− − −

1 1 1 1

1 1

cos sin cosh sinh

2 2

a L a L a L a L

a a

+ +

( ) ( ) ( ) ( ) ( ) ( )2 2

2

1

0.7341 cosh 1.8752 sinh 1.8752 0.7341 cos 1.8752 sin 1.87527.62

0.49217 0.49217

− −= + −

B

( ) ( ) ( ) ( ) ( ) ( )2 2

0.7341 cos 1.8752 sinh 1.8752 0.7341 cosh 1.8752 sin 1.8752

0.24608 0.24608

− −+ −

( ) ( ) ( ) ( )2 20.7341 cos 1.8752 0.7341 cosh 1.8752

0.24608 0.24608

− −− +

( ) ( ) ( ) ( ) ( )2 0.7341 sin 1.8752 sinh 1.8752 cosh 1.8752 sin 1.8752

0.24608 0.24608

−− −

( ) ( ) ( ) ( )cos 1.8752 sinh 1.8752 cos 1.8752 sin 1.8752

0.24608 0.49217− +

( ) ( )cosh 1.8752 sinh 1.8752

0.49217

+


410

2

1 7.6200 11.6374 0.3131 2.0900 6.9730 0.2680 33.2318= + + − − + −B

2

118.1249 12.9395 3.8783 0.5810 21.5948 7.621+ − + − + = B (C.64)

The term 2

17.621B on the right hand side of Equation (C.64) must be equal to L, which

has a value of 7.62. Hence, 2

1B is approximately equal to 1. This implies that 2

1B must

be equal to 1.± In this thesis, 1B has a the value equal of -1. To show that this is true

for any value of n, the author conducts a second example, using the same beam

properties as before, but on this occasion one uses a value n = 3. Using Equations

(C.27b), (C.12b) and (C.59c), the third natural frequency of the beam 3ω is

approximately equal to 482.021 rad/sec, 3a has a value of 1.0308 and C3 has a value

of -0.9992. Inputting these constants into Equation (C.63b) for n = 3 then gives:

2 2 22 2 3 3 3 3 3 3 3 3 33 3

3 3 30


2 2

LC a L a L C a L a L C a L a L

x dx La a a

φ

= + − +

∫ B

2 2 2

3 3 3 3 3 3 3

3 3 3

cosh sin cos coshC a L a L C a L C a L

a a a− − +

3 3 3 3 3 3 3

3 3 3

2 sin sinh cosh sin cos sinhC a L a L a L a L a L a L

a a a− − −

3 3 3 3

3 3

cos sin cosh sinh

2 2

a L a L a L a L

a a

+ +

( ) ( ) ( ) ( ) ( ) ( )2 2

2

3

0.9992 cosh 7.8539 sinh 7.8539 0.9992 cos 7.8539 sin 7.85397.62

2.0614 2.0614

− −= + −

B

( ) ( ) ( ) ( ) ( ) ( )2 2

0.9992 cos 7.8539 sinh 7.8539 0.9992 cosh 7.8539 sin 7.8539

1.0308 1.0308

− −+ −

( ) ( ) ( ) ( )2 20.9992 cos 7.8539 0.9992 cosh 7.8539

1.0308 1.0308

− −− +


411

( ) ( ) ( ) ( ) ( )2 0.9992 sin 7.8539 sinh 7.8539 cosh 7.8539 sin 7.8539

1.0308 1.0308

−− −

( ) ( ) ( ) ( )cos 7.8539 sinh 7.8539 cos 7.8539 sin 7.8539

1.0308 2.0614− +

( ) ( )cosh 7.8539 sinh 7.8539

2.0614

+

2

3 7.620 803494.702 0.000 0.000 1247.677 0.000 1608237.081= + + − − + −B

2

32497.290 1249.614 0.000 0.000 804742.378 7.618+ − + − + = B (C.65)

As before, the term 2

37.618B on the right hand side of Equation (C.65) must be equal

to the length of the beam L, which has a value of 7.62. Therefore, 2

3B has a value

almost equal to 1. This then implies that 2

3B must be equal to 1.± In this thesis, 3B

has a the value equal of -1.


412

Appendix D – Equation for Damping

413

Appendix D

Equation for Damping

D.1 Introduction

In Appendix A, the development of the differential equation of an elastic beam,

included a damping term of 2b

mω without much explanation; however, with the

derivation of the modal equation in Appendix C, one can finally develop this damping

term. Additionally, the numerical damping matrix term C is also derived

D.2 Viscous Damping

Recalling Equation (A.22) the differential equation of an elastic beam is given as:

4 2

4 2( , )

v v vEI m c p x t

tx t

∂ ∂ ∂+ + =

∂∂ ∂ (D.1)

where EI is constant and c is the unknown damping term. In addition, the subscripts

have also been dropped. In order to solve Equation (D.1), one uses the method of

modal superposition; whereby ( ),v x t can be represented as follows

( ) ( ) ( ),n n

n

v x t r t xφ=∑ (D.2)


414

where ( )n xφ is the n-th characteristic shape defined above by Equation (C.12a) and

( )nr t is the n-th function of time which has to be calculated. Substituting Equation

(D.2) into (D.1) gives the following equation:

( ) ( ) ( ) ( ) ( ) ( ) ( )1 1 1

,N N N

iv

n n n n n n

n n n

EI r t x m r t x c r t x p x tφ φ φ= = =

+ + =∑ ∑ ∑&& & (D.3)

where a dot over ( )nr t represents a derivative with respect to time, and a dash over

( )n xφ represents a derivative with respect to x. Substituting Equation (C.7) and

(C.12b) into Equation (D.3) to eliminate the fourth derivative of ( )n xφ gives the

following:

( ) ( ) ( ) ( ) ( ) ( ) ( )4

1 1 1

,N N N

n n n n n n n

n n n

EI r t a x m r t x c r t x p x tφ φ φ= = =

+ + =∑ ∑ ∑&& & (D.4)

Multiplying both sides of Equation (D.4) by ( )i xφ , i = 1, 2, 3, … N, integrating along

the beam length, and using Equation (C.21) and (C.22) gives the following set of

ordinary differential equations governing ( )nr t , n = 1, 2, 3 … N

( ) ( ) ( ) ( ) ( ) ( )4

1 1 1 0

, =1,

LN N N

n n in n in n in i

n n n

EI r t a L m r t L c r t L p x t x dx i Nδ δ δ φ= = =

+ + =∑ ∑ ∑ ∫&& & (D.5)

where 1in


δ = when i n≠ . Rearranging Equation (D.5) in the




415

( ) ( ) ( ) ( ) ( ) ( )4

1 1 1 0

, 1,

LN N N

n in n in n n in i

n n n

mL r t cL r t EIL r t a p x t x dx i Nδ δ δ φ= = =

+ + = =∑ ∑ ∑ ∫&& & (D.6)

Recalling Equation (C.12b), n

a is defined as:

24n na m EIω= (D.7)

Then, substituting Equation (D.7) into (D.6) gives:

( ) ( ) ( ) ( ) ( ) ( )2

1 1 1 0

, 1,

LN N N

n in n in n n in i

n n n

mL r t cL r t mL r t p x t x dx i Nδ δ ω δ φ= = =

+ + = =∑ ∑ ∑ ∫&& & (D.8)

Expanding Equation (D.8) gives the following equation:

( ) ( ) ( ) ( )( )1 1 2 2 3 3 ... i i i N iN

mL r t r t r t r tδ δ δ δ+ + + +&& && && &&

( ) ( ) ( ) ( )( )1 1 2 2 3 3 ... i i i N iN

cL r t r t r t r tδ δ δ δ+ + + + +& & & &

( ) ( ) ( ) ( )( ) ( ) ( ) ( )2 2 2 2

1 1 1 2 2 2 3 3 3

0

... , 1,

L

i i i n N iN imL r t r t r t r t p x t x dx i Nω δ ω δ ω δ ω δ φ+ + + + + = =∫&

(D.9)

Examining the Kronecker Delata function in

δ in Equation (D.9), one finds:

( ) ( ) ( ) ( ) ( )2

1 1 1 1 1

0

,

L

mLr t cLr t m Lr t p x t x dxω φ+ + = ∫&& & at 1i =

( ) ( ) ( ) ( ) ( )2

2 2 2 2 2

0

,

L


( ) ( ) ( ) ( ) ( )2

3 3 3 3 3

0

,

L


or

( ) ( ) ( ) ( ) ( ) ( )2

0

, 1,

L

i i i i imLr t cLr t m Lr t p x t x dx i Nω φ+ + = =∫&& & (D.10)


416

Next, by dividing across by L and putting the right-hand side of Equation (D.10) equal

to zero gives:

( ) ( ) ( )20i i i imr t cr t m r tω+ + =&& & (D.11)

The solution of Equation (D.11) now becomes:

( ) st

ir t Ae= (D.12)

where s and the constants on integration A may be real or complex, depending on

the relative values of the mass, stiffness and damping terms. Substituting Equation

(D.12) into (D.11), one can find the roots of the following equation as:

2 2 0st

ims cs m eω + + = (D.13)

where s defines two roots as follows:

2 2 214

2 2i

cs c m

m mω= − ± −

( )221

22 2

i

cc m

m mω= − ± − (D.14)

From Equation (D.12), one can consider three cases:

( )22

0

2 0

0

ic mω

>

− =<

(overdamped) (D.15a)

(critical damping) (D.15b)

(underdamped) (D.15c)


417

However, only Equation (D.15c) is of interest in this thesis. In the case of

underdamped structures, that is, ( )22 2 0,ic mω − <

the values of s in Equation

(D.14) become the following complex number as:

( ) ( )2 22 21 1

2 22 2 2 2

i i

c cs m c i m c

m m m mω ω = − ± − − = − ± −

( )

( ) ( )

2 2 22

2 2

2

2 22 2

i

i

m cc c ci i

m mm m

ωω

−= − ± = ± − (D.16a)

or

2 2

b i bs iω ω ω= − ± − (D.16b)

and

2b

c

mω= (D.16c)

where b

ω is the damped natural frequency of the beam. Equation (D.16b) can then be

rewritten as follows

2b

c mω= (D.17)

The unknown damping term c in Equation (D.1) is now replaced with Equation

(D.17), which fully derives Equation (A.22) as:

4 2

4 22 ( , )b

v v vEI m m p x t

tx tω

∂ ∂ ∂+ + =

∂∂ ∂ (D.18)

Additionally, Equation (D.16b) can be substituted into (D.12) giving the following:

( ) ( ) ( )2 2exp expi b i b

r t t i tω ω ω= − ± − (D.19)


418

The imaginary term, 2 2 ,i bω ω− in Equation (D.19) can also be defined as:

2

2 2 21 1bi b i i i di

i

ωω ω ω ω ξ ω

ω

− = − = − =

(D.20)

With the aid of Equation (D.16c) and the natural frequency of the system without

damping, i.e. i

k mω = , the damping ratio ξ can be written as follows:

2

22 4 2

b

i

i

c m cc k

m m m k mk

ωξ

ω= = = = (D.21)

Substituting Equation (D.20) and (D.21) into Equation (D.19) gives:

( ) ( ) ( )exp expi i i di

r t t i tξ ω ω= − ±

( ) ( )2exp exp 1i i i i

t i tξ ω ω ξ= − ± − (D.22)

With the aid of Thomson (1993), Equation (D.22) can be written as follows:

( ) ( ) ( )2exp sin 1i i i i

r t X t tξ ω ξ ω ϕ= − − + (D.23)

where the arbitrary constants and X ϕ are determined from the initial conditions.

Equation (D.23) can be simplified using the following general expression:

( )sin sin cos cos sinA B A B A B+ = + (D.24)

resulting in the following equation:


419

( ) ( )( )2 2exp sin 1 cos cos 1 sini i i i i

r t X t t tξ ω ξ ω ϕ ξ ω ϕ= − − ⋅ + − ⋅ (D.25a)

or

( ) ( )2 2sin 1 cos cos 1 sini it

i i ir t e X t X X t X

ξ ω ξ ω ϕ ξ ω ϕ−= − ⋅ + − ⋅ (D.25b)

The effects of Equation (D.25) are shown in Figure D.1.

Figure D.1: Damping oscillation for underdamped motion (Thomson, 1993)

D.3 Numerical damping solution

The damping matrix is dissimilar to the global mass and stiffness matrices because it

cannot be constructed from the element damping matrices during assembly and it

must approximate the overall energy dissipation during the system response (Bathe,

1996). By means of mode superposition and assuming that the damping is

proportional to the vertical velocity of the beam, the normalization of the damping

matrix can be written as follows:

2T

i S i i iω ξ=ψ C ψ (D.26)

where iω is the natural frequency, iξ is the damping ratio of the i-th equation,

eigenvector iψ is the mode shape associated with this eigenvalue and SC is the

itω

( )exp i iX tξ ω−

sinX ϕ

X


420

damping matrix. The mass matrix MS and stiffness matrix KS of the entire structure

are then related to the damping matrix CS using the Rayleigh Method as follows:

0 1S S Sα α= +C M K (D.27)

where 0α and 1α are damping constants. Substituting Equation (D.27) into (D.26):

( )0 1 2T

i S S i i iα α ω ξ+ =ψ M K ψ (D.28)

According to Bathe (1996), when the eigenvector iψ is normalised for the mass and

stiffness matrices, one gets the following equations:

1T

i S i =ψ M ψ (D.29a)

2T

i S i iω=ψ K ψ (D.29b)

By assuming two unequal natural frequencies, then gives:

2

0 1 2i i iα ω α ω ξ+ = (D.30a)

2

0 1 2j j jα ω α ω ξ+ = (D.30b)

Subtracting Equation (D.44b) from (D.44a) then gives:

( ) ( )2 2

1 2i j i i j j

α ω ω ω ξ ω ξ− = − (D.31a)

or

1 2 2

2 2i i j j

i j

ω ξ ω ξα

ω ω

−=

− (D.31b)


421

Substituting Equation (D.31b) into Equation (D.30a) and solving for 0α yields:

3 2 3 2

2

0 2 2 2 2

2 2 2 2 2 22

i i j j i i i j j i i i i j

i i i

i j i j

ω ξ ω ξ ω ξ ω ω ξ ω ξ ω ξ ωα ω ω ξ

ω ω ω ω

− − + + −= − + = − −

( )2 2

2 2 2 2

22 2 i j i j j ji j j i i j

i j i j

ω ω ω ξ ω ξω ω ξ ω ξ ω

ω ω ω ω

−−= =

− − (D.32)

Should it be considered that i j

ξ ξ ξ= = , then Equation (D.31b) and (D.32) can further

reduced to the following:

( ) ( )( ) ( )0 2 2

2 2 2i j i j i j i j i j

i j i ji j i j

ω ω ω ξ ω ξ ω ω ξ ω ω ω ω ξα

ω ω ω ωω ω ω ω

− −= = =

− ++ − (D.33a)

( ) ( )( )( )1 2 2

2 2 2i j i j

i j i ji j i j

ξ ω ω ξ ω ω ξα

ω ω ω ωω ω ω ω

− −= = =

− ++ − (D.33b)

Rearranging Equation (D.30a) with subscripts i and j dropped, the relationship

between the damping coefficient ξ and the natural frequencies ω for Rayleigh

damping can be seen in Figure D.2.

0

1

2

3

4

5

6

7

8

0 1 2 3 4 5 6 7 8

Mass Proportional

Stif fness Proportional

Combined

Figure D.2: Damping coefficient versus natural frequencies for Rayleigh damping

ω

ξ

0

01, when 0

2

αξ α

ω= =

10, when 0

2

α ωξ α= =

0 1

2 2

α α ωξ

ω= +


422

0 1

2 2

α ωαξ

ω= + (D.34)

The equation of equilibrium governing the linear dynamic response of a structure,

including damping, is now written as follows:

S S S S S S S+ +M U C U K U = F&& & (D.35)

where MS, CS and KS are the global mass, damping and stiffness matrices; FS is the

force vector; , and S S S

U U U& && are the displacement, velocity and acceleration vectors

of the entire finite element assemblage.

Appendix E –Newmark Time Integration Scheme

423

Appendix E

Newmark Time Integration Scheme

E.1 Introduction

In this appendix, the author derives the Newmark time integration scheme as this

system is used extensively throughout this thesis to solve the finite element equations.

E.2 Newmark Time Integration

The Newmark time integration scheme is an implicit scheme because it solves the

equation of motion at time ,t t+ ∆ by directly combining the displacements, velocities

and acceleration equations at time ;t t+ ∆ whereas, an explicit scheme solves the

equation of motion at time ,t t+ ∆ by combining the displacements, velocities and

acceleration equations at time .t Recalling Equation (D.47), the equation of motion is

defined as follows:

MU + CU + KU = F&& & (E.1)

where M, C and K are the mass, damping and stiffness matrices, respectively, F is the

external force vector, and , and U U U& && are the displacement, velocity and acceleration

vectors of the finite element assembly. Therefore, at time t, Equation (E.1) becomes


424

t t t tMU + CU + KU = F&& & (E.2)

It is assumed that at time t, the values of the displacement, velocity and acceleration

are known, while at the end of the time interval t t+ ∆ these values are unknown.

With the aid of Taylor series, the displacement and velocity at time t t+ ∆ can be

written as follows:

2 3

. . .2 6

t t t t t t

t tt+∆

∆ ∆= + ∆ + + +U U U U U& && &&& (E.3a)

2

. . .2

t t t t t

tt+∆

∆= + ∆ + +U U U U& & && &&& (E.3b)

Newmark (1952) then condensed these equations to the following form by

introducing the parameters β and γ:

23

2t t t t t t

tt tβ+∆

∆= + ∆ + + ∆U U U U U& && &&& (E.4a)

2

t t t t tt tγ+∆ = + ∆ + ∆U U U U& & && &&& (E.4b)

By assuming that the acceleration remains linear within the time step, one gets the

following equation:

( )t t t

tt

+∆ −=

∆

U UU

&& &&

&&& (E.5)

Substituting Equation (E.5) into Equation (E.4a) and (E.4b) gives the displacement

and velocity at time t t+ ∆ as defined by Newmark (1959) as follows:


425

( )23

2

t t t

t t t t t

tt t

tβ

+∆

+∆

−∆= + ∆ + + ∆

∆

U UU U U U

&& &&

& &&

22 2

2t t t t t t

tt t tβ β+∆

∆= + ∆ + + ∆ − ∆U U U U U& && && &&

21

2t t t t t

t tβ β +∆

= + ∆ + − + ∆

U U U U& && && (E.6a)

( )2

t t t

t t t tt t

tγ

+∆

+∆

−= + ∆ + ∆

∆

U UU U U

&& &&

& & &&

( ) t t t t t

t tγ +∆= + ∆ + ∆ −U U U U& && && &&

( )1t t t t

tγ γ +∆ = + − + ∆ U U U& && && (E.6b)

Examining Equation (E.6), the acceleration term at the end of the time interval t t+∆U&& ,

despite being unknown at this time, can be related to the displacement and velocity at

time .t t+ ∆ By rearranging Equation (E.6a), the acceleration at time t t+ ∆ can be

written as:

( )2

1 1 11

2t t t t t t t

t tβ β β+ ∆ + ∆

= − − − −

∆ ∆ U U U U U&& & && (E.7)

Substituting Equation (E.7) into (E.6b), then gives the velocity at time t t+ ∆ as:

( ) ( )2

1 1 11 1

2t t t t t t t t t

tt t

γ γβ β β

+∆ +∆

= + − + − − − − ∆

∆ ∆ U U U U U U U& & && & &&

( ) ( )12

t t t t t t tt tt

γ γ γγ γ

β β β+∆

= + − ∆ + − − − − ∆

∆ U U U U U U& && & &&

( ) 1 12

t t t t tt

t

γ γ γ

β β β+ ∆

= − + − − ∆ −

∆ U U U U& && (E.8)


426

At time t t+ ∆ Equation (E.2) now becomes:

t t t t t t t t+∆ +∆ +∆ +∆MU + CU + KU = F&& & (E.9)

Substituting Equation (E.7) and (E.8) into Equation (E.9) gives:

2

1t t

t t

γ

β β+ ∆

+ + =

∆ ∆ M C K U

2

1 1 11

2t t t t t

t tβ β β+∆

+ + + −

∆ ∆ F M U U U& &&

1 12

t t ttt

γ γ γ

β β β

+ + − + − ∆

∆ C U U U& && (E.10)

Solving Equation (E.10), one finds the displacement value t t+∆U at time t t+ ∆ . This

known value can then be substituted into Equation (E.7) and (E.8) to obtain the

acceleration and velocity at time ,t t+ ∆ in accordance with ANSYS Theory

Reference (2002).

Alternatively, one can assume a value for the acceleration t t+∆U&& at the end of the time

interval in order to compute the velocity t t+∆U& and

t t+∆U using Equation (E.6b) and

(E.6a). The computed displacement t t+∆U is then used to compute the resisting force

for the structure. Next the applied loads and resisting forces t t+∆F at the end of the

interval are used to recompute the acceleration t t+∆U&& at the end of the time interval.

The derived acceleration is then compared with assumed acceleration. If these values

are the same, the calculation is complete. If the values are different, then the

calculation is repeated using a different assumed acceleration (Newmark, 1959).


427

E.3 Defining the parameter γ and β

According to Newmark (1959), the term γ must have a value of 1 2 , to avoid

damping the results; therefore, one can rewrite Equation (E.6b) as follows:

2 2

t t tt t t t+∆+∆

= + + ∆

U UU U

&& &&& & (E.11)

In Figure E.1, the author plots the relationship of three β values on the acceleration

between tU&& and

t t+∆U&& over the time interval t∆ . Firstly, when β has a value of 1 6,

the acceleration over the time interval t∆ remains linear. Secondly, one can see that

when β has a value of 1 4, a uniform value of acceleration occurs during the time

interval equal to the mean of the initial and final values of acceleration, and thirdly,

when β has a value of 1 8, a step function with a uniform value equal to the initial

value of the first half of the time interval and a uniform value equal to the final value

for the second half of the time interval can be seen. In ANSYS, the default setting for

γ is 1 2 and β is 1 4 . These parameters determine the accuracy and stability of the

scheme and are unconditionally stable at these particular values.

Figure E.1: Plot of acceleration between time t and t t+ ∆ for three values of β

t t t+ ∆

time

Acc

eler

atio

n

t∆

t t+∆U&&

tU&&

1 8β =

1 6β =

1 4β =


428

Appendix F - Boyne Viaduct - Technical Parameters

429

Appendix F

Boyne Viaduct – Technical Parameters

F.1 Dimensions & Geometrical Properties

In this study, one is primarily interested in the centre span of the Boyne Viaduct

railway bridge; hence, all dimensions, sections sizes and bridge properties related to

this particular bridge structure are present in this section. Figure F.1 presents a side

view, while a sectional view of the bridge structure can be seen in Figure F.2, along

with all critical dimensions.

Figure F.1: Front elevation of Boyne Viaduct railway-bridge

Bridge Length = 80.77m

Number of spans = 10

Distance between main nodes of each span = 8.077m

Distance between the upper and lower chords at the supports = 5.80m

Distance between the upper and lower chord at midspan = 9.14m

All other heights along the bridge = ( ) ( )9.14 5.80 sin 5.80xL

π− ⋅ +

Width between truss = 5.5m

Track gauge = 1.5m


430

Figure F.2: Sectional view of Boyne Viaduct railway-bridge

A three-dimensional representation of the Boyne Viaduct railway bridge is presented

in Figure F.3, which only takes into account the main nodes of the truss. One can

observe from the diagram, two longitudinal beams running parallel with truss. The

main purpose of these beams is to transfer the weight of the train safely to cross

beams (located perpendicular to the longitudinal beams at every 8.077m), which in

turn transfers the weight to the truss.

Figure F.3: Three-dimensional view of Boyne Viaduct railway-bridge

In Table F.1 the bridge fixities of the three-dimensional model are defined.

Support 1

Support 3

Support 5

Support 7

Support 2

Support 4

Support 6

Support 8

Longitudinal beams

Rail Rail


431

Table F.1 Support Fixities

Support X Y Z Rot X Rot Y Rot Z

1 Fixed Fixed Fixed Free Free Free

2 Free Fixed Fixed Fixed Fixed Free

3 Free Fixed Free Free Free Free




7 Fixed Fixed Fixed Free Free Free

8 Free Fixed Fixed Fixed Fixed Free

Next, using the geometrical dimensions of a section, one can evaluate some sectional

properties of typical members of the truss. In each case, the sectional properties

related to that particular section are given in Table F.2. A cross-sectional view of the

upper chord of the bridge is shown in Figure F.4. It should be noted that no sectional

views are available for the lower chord; therefore, it is assumed that the properties of

the lower chord are similar to those of the upper chord of the truss.

Figure F.4: Cross-sectional view of the upper chord of the truss

Next, one examines a typical vertical member located along the bridge as illustrated in

Figure F.5. It is assumed that all vertical members are of a similar shape and size.


432

Figure F.5: Cross-sectional view of a vertical member

A typical diagonal member located along the bridge is illustrated in Figure F.6. Again

it is assumed that all diagonal members are of a similar shape and size.

Figure F.6: Cross-sectional view of a diagonal member on bridge

Next, one considers the longitudinal beams that are located beneath the bridge

decking and run parallel to the track. Figure F.7 shows a typical cross-sectional view

of the longitudinal beam

Figure F.7: Cross-sectional view of the longitudinal beam on bridge


433

Finally, a cross-sectional view of a typical cross beams is presented in Figure F.8.

Figure F.8: Cross-sectional view through the cross beam on bridge

Table F.2 Section properties of key bridge members

Cro

ss-S

ecti

on

Are

a (m

2)

Mas

s p

er m

etre

(t/

m3)

– a

ll s

teel

pro

per

ties

Sec

on

d m

om

ent

of

iner

tia

abou

t Y

-Y (

m4)

Sec

on

d m

om

ent

of

iner

tia

abou

t Z

-Z (

m4)

Th

ick

nes

s in

Y (

m)

Th

ick

nes

s in

Z (

m)

Upper & Lower Chords 0.05 0.393 - - 0.65 0.76

Vertical Members 0.025 0.196 - - 0.45 0.45

Diagonal Members 0.012 0.094 - - 0.3 0.3

Longitudinal Beams 0.032 0.251 0.00014 0.0034 0.91 0.4

Cross Beams 0.044 0.345 0.00021 0.0042 0.94 0.46


434

F.2 Static analysis of an unloaded Boyne Viaduct

Table F.3 presents results obtained from a static analysis of an unloaded Boyne

Viaduct railway bridge using the dimensions and geometrical properties of Section

F.1. These results include the vertical displacement, axial force in the upper and lower

chords and bending moment at mid-span of the bridge as well as its overall weight. It

should be noted that the geometrical properties of the simply supported beam are

defined in Section F.3.

Table F.3: Static analysis of the Boyne Bridge

Sim

ply

su

pp

ort

ed b

eam

Tw

o-d

imen

sio

na

l tr

uss

Th

ree-

dim

ensi

on

al

bri

dg

e

Maximum deflection at mid-span (m) = 0.0256 0.0256 0.0300

Total weight of the bridge structure (kN) = 933.0 933.0 2527.4

Maximum axial force in the upper chord at midspan (kN) = N/A -1059 -1396.8

Maximum axial force in the lower chord at midspan (kN) = N/A 1052.8 904.8

Maximum bending moment at midspan (kNm) = 9420 N/A 104.6*

* = longitudinal beam of the three-dimensional bridge

F.3 Representing a truss as a simply supported beam

In Chapter 5, the author analysis the Boyne Viaduct as a simply supported beam using

the developed modal model; therefore, this sub-section examines Fryba’s (1999)

concept of modelling a truss railway bridge as a simply supported beam. In the

previous section, one found that the overall weight of the two-dimensional truss is 933


435

kN, thus, the mass per unit length m of the simply supported beam can be calculated

as follows:

Gm

gL=

933.0

80.77 9.81=

× 31.1775 t/m= (F.1)

where G is the total weight of the bridge, g is gravity, and L is the length of the beam.

It is also found that the static deflection at mid-span of the two dimensional truss is

equal to 0.0253 m. Therefore, using the equation for the deflection of a simply

supported beam subjected to a uniformly disturbed load as given in Equation (F.2),

one can compute an appropriate second moment of inertia I for the beam.

45

384st

mgLv

EI= (F.2)

where vst is the static deflection at mid-span of the beam, and E is Young’s Modulus

of Elasticity of the beam. Rearranging Equation (F.2) in terms of I gives:

45

384st

mgLI

Ev= (F.3)

Assuming E is 2.05 x 108 kN/m

2, then using Equation (F.3), I is 1.22 m

4. Since

ANSYS requires the user to specify the cross-section area of two and three-

dimensional beam elements, one assumes that the density of the steel used in the

simply supported beam ρ is 7.85 t/m3; therefore, the cross-section area A = m/ρ, which

is equal to 0.15 m2.


436

F.4 Modal analysis of the Boyne Viaduct railway bridge

The first 5 natural frequencies (measured in Hertz) for the Boyne Viaduct are given in

Table F.4. These results include an analytical modal solution using Equation (C.48),

and three finite element modal solutions with the bridge represented as a simply

supported beam, two-dimensional truss and three-dimensional.

Table F.4: Natural frequencies of the Boyne Viaduct.

Natural frequencies in the y-direction (Hz)

j = 1 j = 2 j = 3 j = 4 j = 5

Simply supported beam (analytical) 3.51 14.04 31.58 56.14 87.73

Simply supported beam 3.49 13.70 29.97 51.32 76.72

Two-dimensional truss 3.40 9.87 15.97 21.01 28.78

Three-dimensional bridge 3.12 8.72 13.86 18.28 24.99

In Figure F.9, F.10 and F.11, the author plots the first 5 mode shapes of the simply

supported beam, two-dimensional truss and three-dimensional truss, respectively.

(a) 1st mode – 3.49 Hz (b) 2

nd mode – 13.70 Hz

(c) 3rd

mode – 29.97 Hz (d) 4th

mode – 51.32 Hz

(e) 5th

mode – 76.72 Hz

Figure F.9 First 5 modes shapes in the y-direction for the simply supported beam


437

(a) 1st mode – 3.40 Hz (b) 2

nd mode – 9.87 Hz

(c) 3rd

mode – 15.97 Hz (d) 4th

mode – 21.01 Hz

(e) 5th

mode – 28.78 Hz

Figure F.10 First 5 modes shapes in the y-direction for the two-dimensional truss

(a) 1st mode – 3.12 Hz (b) 2

nd mode – 8.72 Hz

(c) 3rd

mode – 13.86 Hz (d) 4th

mode – 18.28 Hz


438

(e) 5th

mode – 24.99 Hz

Figure F.11 First 5 modes shapes in the y-direction for the three-dimensional truss

Appendix G – Railway Vehicle Dynamics

439

Appendix G

Railway Vehicle Dynamics

G.1 Introduction

Since this thesis involves the dynamic response of vehicle traversing a railway bridge,

it is important that each railway vehicle is simulated correctly. In Chapter 3, the

author is primarily interested in modelling each railway vehicle as a series of moving

forces; whereas, in Chapter 4, 5 and 6 both a two-dimensional and three-dimensional

railway vehicle, comprising of a vehicle body, bogies and axles separated by primary

and secondary suspensions, is sought. The railway vehicles adopted in this study are

the 201 Class Irish-Rail locomotive (six-axles) as well as the Mark3 railway coach

(four axles). A photo of the 201 Class locomotive and Mark3 railway coach is shown

in Figure G.1, respectively. According to Wikipedia (2008), the 201 Class locomotive

is the most powerful diesel locomotives operating in Ireland since 1994. It has an

overall weight of approximately 111.5 tonnes and a maximum speed of 164 km/h

(102 mph). The Mark3 railway coach is a passenger carriage and has an overall

weight of approximately 48 tonnes.


440

Figure G.1: Photos of Irish-Rail railway vehicles: (a) 201 Class Irish-Rail

locomotive; (b) Mark3 railway coach

G.2 Axle spacing and weights as a moving force

The axle spacing and weights of the 201 Class locomotive and Mark3 railway coach

were supplied by the Irish Rail’s Structural Design Office, Inchicore, Dublin and are

presented in Figure G.2. As illustrated in Figure G.2, each wheel of the locomotive

and railway coach exerts a force of 91.25 kN on the rail and each wheel of the railway

coach exerts a force of 58.86 kN on the rail, respectively. These values are computed

as follows:

(a)

(b)


441

Wheel load per track Overall weight of the vehicle in tonnes gravity

Number of axles two wheels per axle

×=

×

Locomotive wheel load 111.5 9.81

= 91.25 kN6 2

×=

×

Railway coach wheel load 48 9.81

= 58.86 kN4 2

×=

×

Additionally, to ensure that the Boyne Viaduct railway bridge experiences the full

weight of the train over its entire length, the author has chosen that each train consists

of a single six-axle locomotive, 201 Class type, followed by three four-axle Mark3

railway coaches such that the total distance between the front axle and rear axle of the

train is equal to ( ) ( )21.051 3 23.000 1.969 2.200 85.882 m.+ × − + = Examining Figure

G.2, the reader can also see that the distance between two axles of a single bogie Aw

are 1.689 m and 2.019 m for the 201 Class locomotive and 2.600 m for the Mark3

railway coaches. In addition, the distance between the rear axle of the locomotive and

the front axle of the railway carriage Cw is 1.969 2.200 4.169 m,+ = and the distance

between the rear axle of a carriage and front axle of the carriage behind it Cw is

2.200 2.200 4.400 m.+ = Both of these lengths, Aw and Cw, are clearly shorter than

the distance between the rear axle of front bogie and the front axle of the rear bogie

(Bw–Aw), which is 9.697 for the locomotive and 13.400 m for the railway carriage.

Hence, one can state that ( ).w w w w

A C B A< < −

(a)

201 Class locomotive

Gross Weight = 111.5 tons

Axle Weight = 18.6 tons




442

Figure G.2: Typical axle spacing and load of Irish-Rail vehicles: (a) 201 Class

locomotive; (b) Mark 3 railway coach

G.3 Axle positioning for maximum loading of Boyne Bridge

Throughout this thesis, the author has been dividing the dynamic results (deflection

and axial forces) by the maximum static results at mid-span of the loaded Boyne

Viaduct Railway Bridge to compute the dynamic amplification factor DAFU and

DAFA. The maximum static results (deflection and axial forces) at mid-span of the

Boyne Bridge are determined from the position of the front axle of the train. For a

single moving load, the axle is positioned at 80.77/2 = 40.385 m from the left hand

support, while for multiple moving forces the front axle is positioned at 60.577 m

from the left hand support, for the maximum static load, as shown in Figure G.3.

Figure G.3: Axle positioning that causes the maximum static results of Boyne

Bridge: (a) single moving load; (b) multiple moving forces

(b)

Mark 3 Railway Coach

Gross Weight = 48 tons

Axle Weight = 12 tons



(a)

(b)

40.385 m

60.577 m


443

Figure G.3b shows that the maximum static results (deflection and axial forces) occur

when the rear wheel of the locomotive is positioned slightly to the right of the mid-

span of the Boyne Bridge, mainly because the locomotive is 130% heavier than a

single railway carriage. The maximum static deflections at mid-span of the loaded

Boyne Bridge used to compute DAFU are:

Deflection of 2D bridge due to a single moving load (used in Figure 3.27a) = 40.91 mm

Deflection of 2D bridge due to multiple moving loads (used in Figure 3.39) = 31.08 mm

Deflection of 3D bridge due to multiple moving loads (used in Figure 3.45a) = 27.66 mm

One can see that the maximum static deflection of the two-dimensional bridge for the

single moving point force, which has a P/G ratio of 1, is approximately 30% greater

than the maximum static deflection of the same bridge for the multiple moving forces.

Nonetheless, the author shows with the aid of Figure G.4 that the dynamic

amplification factor remains unchanged even when the P/G ratio is reduced to 0.7.

1.00

1.05

1.10

1.15

1.20

0 50 100 150 200 250 300

Speed (km/hr)

DA

FU

Multiple Moving Loads

Single Moving Load - P/G = 1.00 - Figure 3.28

Single Moving Load - P/G = 0.7

Figure G.4: Dynamic amplification factor unchanged even when the P/G ratio for a

single moving load is reduced from 1.0 to 0.7


444

The maximum static deflection of the loaded two-dimensional Boyne Bridge, for a

single moving point force with a P/G ratio of 0.7, is equal to 28.64 mm. The

maximum static axial forces at mid-span of the loaded Boyne Bridge used to compute

DAFA are as follows:

Bottom chord of 2D bridge for the single moving load (used in Figure 3.27b) = 1834 kN

Top chord of 2D bridge for the single moving load (used in Figure 3.27b) = -1909 kN

Bottom chord of 2D bridge for the multiple moving loads (used in Figure 3.40b) = 1346 kN

Top chord of 2D bridge for the multiple moving loads (used in Figure 3.40b) = -1369 kN

Bottom chord of 3D bridge for the multiple moving loads (used in Figure 3.45b) = 871 kN

Top chord of 3D bridge for the multiple moving loads (used in Figure 3.45c) = -1350 kN

G.4 Exact model with overlapping time functions

In Section 3.3.1.2, the author presents the simple model with overlapping time

functions, which can be seen again in Figure G.5. This figure shows the spatial and

time domain of two moving forces, with the same magnitude, traversing a beam; for

several values of wA l for node I. For completeness, the author will now plot the time

domains, of the two moving forces using the exact model with overlapping time

functions (forces and moment) as illustrated in Figure G.6.


2wA l=

1Q1P

l

i

P

4l

c

t0

(1)

(2)

(3)

(4)

(5)

(6)


445




Figure G.5: Relationship between spatial and time domain under different load

conditions for two moving forces using the simple model

7

2

l

c

t

P

1Q 1P

1.5wA l=

l

i

3l

cl

wA l=

1Q1P

i t

P

t

0

0

0



(1)

(2)

(3)

(5)

(6)

(4)

(1)

(2)

(6)

(6) 5

2

l

cl

0.5wA l=P

i

1Q 1P

(1)

(2)

(3)

(5)

(4)

(3) (4)

(5)


446





Figure G.6: Time domain of the two moving forces using the exact model with

overlapping time functions

P

P



( )1

( )2

( )3 ( )6( )4

( )5

( )1

( )2

( )3 ( )6

( )4

( )5

( )1

( )2

( )3 ( )6( )4

( )5

( )4

( )3( )2

( )6

( )1

( )5

( )1

( )2

( )3 ( )6( )4

( )5P

P

t

( )4

( )3( )2

( )6

( )1

( )5

( )4

( )3( )2

( )6

( )1

( )5

t

t

t

( )6

( )1 t

t

t

t

P

P M

M

M

M


447

G.5 Vehicle dimensions and parameters

As shown in Figure G.7, each vehicle consists of a vehicle body supported by a pair

of bogies, with each bogie supported by axles. Finally a pair of wheels supports each

axle. The bogies are connected to the axles through primary suspensions and to the

vehicle body through secondary suspensions, with each suspension consisting of a

spring and dashpot. In the two-dimensional models, one assumes that the weight of all

components is halved.


primary suspension

Hertian spring

vehicle body

bogie

wheelset

vehicle body

bogie

Side View

bogie

Front View

wheels wheels


primary suspension

Hertian spring

vehicle body

bogie

wheelset

vehicle body

bogie

Side View

bogie

Front View

wheels wheels

Figure G.7: Three-dimensional railway carriage model (a) descriptive; (b) using

symbolization

(a)

2 yk 2 y

c

1yk 1y

k

2 yk

1yc

1yc

2 yc

HkH

kw

Mw

Mw

Mw

M

,v zzv

M I

,b zzb

M I

(b)

,b zzb

M I

,v xxv

M I

bM

y

x

1 1,z z

k c

1 1,y y

k c

1 1,z z

k c

1 1,y y

k c

Front View

Side View

4wA

3wA 2w

A1w

Aw

b

vb

wB

y

z

vl

0h

1h

3h


448

The total mass of each wheel, bogie frame and vehicle body is denoted by the

symbols Mw, Mb and Mv. The primary spring stiffness and damping is given by k1 and

c1, while k2 and c2 denote the secondary spring stiffness and damping. The Hertzian

spring stiffness, also known as the wheel-rail contact stiffness, is given by kH.

Longitudinal and lateral springs are required between the vehicle body and the bogie

and also between the bogie and axles to prevent the structure from becoming a

mechanism. These horizontal springs are independent from the main suspension

springs.

Each wheel of the train is modelled using lumped masses, while elastic beam

elements are used to model the bogie frame and vehicle body. The primary and

secondary suspensions are then modelled using spring elements. The author’s six-axle

201 Class locomotive and four-axle Mark3 railway coach modelled in ANSYS can be

seen in Figure G.8.

Figure G.8: Six-axle 201 Class locomotive and four-axle Mark3 railway coach

modelled using beam, spring and lumped mass elements in ANSYS

(Bowe and Mullarkey, 2005)

V1

W1 – First wheel of the train (node)

W7 – Seventh wheel of the train (node)

V1 – Vehicle body of the locomotive (node)

V2 – Vehicle body of the railway coach (node) V2

W1

W7


449

The author indicates in Figure G.8, in particular, the 1st (W1) and 7

th (W7) wheel of

the train, one is the front wheel of the locomotive and the other is the front wheel of

the first railway coach. The nodes V1 and V2 on the vehicle bodies of the locomotive

and railway coach, respectively, are also shown. These nodes are examined in the

results.

A list of vehicle properties and dimensions for both the Class 201 locomotive and

Mark3 railway coach adopted in this study are presented in Table G.1. The symbol X

marked in the left hand column indicates that this particular value is not used or

required in the two-dimensional vehicle model.

Table G.1: Vehicle dimensions and parameters

Data Symbols Unit Loco 201 MK3

coach

Mass properties

Mass of vehicle body Mv t 64.48 35.74

x Roll inertia of vehicle body Ixxv tm2 123.0 68.2

x Yaw inertia of vehicle body Iyyv tm2 2008 1287

Pitch inertia of vehicle body Izzv tm2 2002 1284

Mass of bogie frame Mb t 10.18 3.15

x Roll inertia of bogie frame Ixxb tm2 10.18 3.15

x Yaw inertia of bogie frame Iyyb tm2 21.73 4.91

Pitch inertia of bogie frame Izzb tm2 11.55 1.76

Mass of wheel Mw t 4.52 1.50

Roll inertia of wheel Ixxw tm2 4.52 1.50

Suspension stiffness

Primary suspension in the longitudinal direction k1x kN/m 4240 20260

Secondary suspension in the longitudinal direction k2x kN/m 320 422

Primary suspension in the vertical direction k1y kN/m 1470 3185

Secondary suspension in the vertical direction k2y kN/m 630 566

x Primary suspension in the lateral direction k1z kN/m 2120 10130

x Secondary suspension in the lateral direction k2z kN/m 160 211

Suspension damping

Primary suspension in the longitudinal direction c1x kNs/m 1.00 0.000

Secondary suspension in the longitudinal direction c2x kNs/m 32.00 41.44

Primary suspension in the vertical direction c1y kNs/m 4.00 32.41

Secondary suspension in the vertical direction c2y kNs/m 20.00 26.24

x Primary suspension in the lateral direction c1z kNs/m 1.00 0.00

x Secondary suspension in the lateral direction c2z kNs/m 32.00 41.44


450

Dimension - longitudinal direction - x

1st axle to c.o.g. of bogie frame Aw1 m 1.689 1.300

2nd axle to c.o.g. of bogie frame Aw2 m 0.000 1.300

3rd axle to c.o.g. of bogie frame Aw3 m 2.019 1.300

4th axle to c.o.g. of bogie frame Aw4 m 1.689 1.300



Rear axle of a vehicle to front axle of next vehicle Cw m 4.169 4.400

Front bogie frame c.o.g. to c.o.g. of rear bogie frame Bw m 13.405 16.000

Half of bogie frame to c.o.g. of vehicle body lbg m 6.703 8.000

Overall length of vehicle body lv m 19.113 20.600

Dimension - vertical direction - y

Height from rail to wheel centre h0 m 0.508 0.460

Height from rail to c.o.g of bogie h1 m 0.843 0.600

Height from rail to secondary suspension h2 m 0.843 0.600

Height from rail to c.o.g. of vehicle body h3 m 1.393 0.220

Overall height of vehicle body hv m 2.200 2.200

Dimension - lateral direction - z

x Gauge width of wheelset bw m 1.500 1.500

x Overall width of vehicle body bv m 2.000 2.000

Vehicles components such as the bogies and vehicle body are represented by elastic

beam elements within the ANSYS finite element program, as this system is easy

implemented. However, this technique required the author to pre-calculate the centre

of gravity of each component and its mass moment of inertia. Additionally, these

values can change if the geometry, dimensions or number of elements in the model is

altered. This is illustrated in the following example, where the centre of gravity and

mass moments of inertia (roll, yaw and pitch) of the vehicle body of the Class 201

locomotive is computed. A three-dimensional representation of the locomotive

modelled in ANSYS using beam elements can be seen in Figure G.9. It should be

noted in this particular example that the author has assumed that the vertical and

lateral beam elements are insignificant compared with the longitudinal beams; hence

are omitted in the calculations. Additionally, one assumes that the bottom beam

contribute 75% of the total mass of the vehicle body, while the top beam only

contribute 25%.


451

Figure G.9: Three-dimensional locomotive in ANSYS using beam elements

Next, the author isolates the vehicle body of the locomotive, which is shown in Figure

G.10. Examining this diagram, the reader can see that the vehicle body comprises

eight beam elements; six beam elements on the bottom surfaces, while only two beam

elements on the top surface, numbered from 1 to 8 as indicated (model A).

Figure G.10: Isolated vehicle body modelled using 8 beam elements (model A)

Due to symmetry, the centre of gravity in the longitudinal and lateral direction of the

vehicle body is located at a distance of 9.5565 m and 1.000 m from the origin,

respectively. However, in order to calculate the centre of gravity of the vehicle body

in the vertical direction, the author must use the following equation (ANSYS, 2002):

1

N

c vi vi v

i

Y m y M=

=∑ (G.1)


452

where area density lengthvi

m = × × of the i-th beam element, Mv is the total mass of

the vehicle body and vi

y is the distance from neutral axis of the beam element to the

origin of the vehicle body i.e. the intersection of the x, y and z-axis. With the aid of

Table G.1 and Figure G.10, the author computes the vertical centre of gravity of the

vehicle body as follows:

1 1 2 2 3 3 8 8 . . . +c v v v v v v v v v

Y m y m y m y m y M= + + +

( ) ( ) ( ) ( ) 3.402 0 17.376 0 3.402 0 +8.06 2.2 2 64.48= + + ×

35.464 64.48 0.550 m= =

Using the three centres of gravity of the vehicle body i.e. 9.5565 m, 0.550 m and

1.000 m in the longitudinal, vertical and lateral directions, respectively, one now

calculates the three equivalent mass moments of inertia (roll, yaw and pitch) for the

vehicle body using the following equations (ANSYS, 2002):

( ) ( )2 2

1

N

xx vi vi vi

i

I m y z=

= + ∑ (rolling inertia) (G.2a)

( ) ( )2 2

1

N

yy vi vi vi

i

I m x z=

= + ∑ (yawing inertia) (G.2b)

( ) ( )2 2

1

N

zz vi vi vi

i

I m x y=

= + ∑ (pitching inertia) (G.2c)

where vi

x , vi

y and vi

z is the distance from the neutral axis of the beam element at its

own local centre of gravity to the centre of gravity of the vehicle body. With the aid of

Figure G.10 and Equation (G.2), the mass moments of inertia of the vehicle body are

calculated as follows:


453

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

( )

2 2 2 2 2 2

1 1 1 2 2 2 8 8 8

2 2 2 2 2 2

2 2

2

. . . +

3.402 0.55 1.00 17.376 0.55 1 3.402 0.55 1.00

+8.06 1.65 1.00 2

4.431+22.632+4.431+30.003 2 123.0 tm

xx v v v v v v v v vI m y z m y z m y z = + + + + +

= + + + + +

+ ×

= × =

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

( )

2 2 2 2 2 2

1 1 1 2 2 2 8 8 8

2 2 2 2 2

2

2

. . . +

3.402 8.212 1.00 17.376 0 1.00 3.402 8.212 1.00

+8.06 0 1.00 2

232.822+17.376+232.822+8.060 2 982.2 tm

yy v v v v v v v v vI m x z m x z m x z = + + + + +

= + + + + +

+ ×

= × =

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

( )

2 2 2 2 2 2

1 1 1 2 2 2 8 8 8

2 2 2 2 2

2

2

. . . +

3.402 8.212 0.55 17.376 0 0.55 3.402 8.212 0.55

+8.06 0 1.65 2

230.450+5.256+230.450+21.943 2 976.2 tm

zz v v v v v v v v vI m x y m x y m x y = + + + + +

= + + + + +

+ ×

= × =

However, it should also be noted that these above values are dependent on the

quantity and positioning of beam elements in the locomotive model shown in Figure

G.10. Supposing the bottom surface of the vehicle body is now discretized into four

beam elements instead of three, such that the middle element is divided into two

smaller elements as shown in Figure G.11, then the three equivalent mass moments of

inertia using Equation (G.2) are equal to:

2123.0 tmxx

I = (model B shown in Figure G.11)

21391.9 tmyyI = (model B shown in Figure G.11)

21385.9 tmzz

I = (model B shown in Figure G.11)


454

In a similar manner, if the vehicle body is also discretized into several beam elements

i.e. ten elements on the top and bottom surface, then the equivalent mass moment of

inertia will again change to the following values:

2123.0 tmxx

I = (model C shown in Figure G.12)

22002 tmyyI = (model C shown in Figure G.12)

22005 tmzz

I = (model C shown in Figure G.12)

Figure G.11: Isolated vehicle body modelled using 10 beam elements (model B)

Figure G.12: Isolated vehicle body modelled using 40 beam elements (model C)


455

Beyond ten beam elements used to simulate the vehicle body, the additional increase

in the mass moment of the inertia of the vehicle body becomes less significant.

Thus far, it has been shown that the equivalent mass moment of inertia of the vehicle

body tends to increase as the number of beam elements in the model increases.

However, in the following example it will be shown that this additional mass moment

of inertia in the three different simulation i.e. model A, B and C has little or no effect

on the finite element solution. The reasoning been is that the mass matrix of the beam

element tends to distribute the mass to both ends of the beam whereas the equivalent

mass moment of inertia tends to assume that the mass is located at the centre of

gravity of the beam. Therefore, one can confidently simulate the vehicle body with

fewer beam elements in the finite element solution, while still achieving the same

equivalent mass moment of inertia of a vehicle body with many beam elements.

By assuming the three different vehicle bodies i.e. model A, B and C are supported by

vertical, longitudinal and lateral suspension springs located at its free ends along the

bottom surface and that massless beam elements support the top surface beams as

shown in Figure G.9, the author is now in a position to conduct a transient analysis on

each vehicle body. Firstly, one examines the effects of releasing the vehicle body

from its unsettled static position such that it will undergo free-vibration in the

transient analysis, while secondly an additional rotational displacements (about the x-

axis and z-axis) are added to the vehicle body in the static analysis; thus, causing the

vehicle body to pitch and roll as well as free-vibrate in the transient analysis.

The vertical displacement of each of the vehicle bodies undergoing free vibration only

is presented G.13. It can be seen from the results that the different mass moment of


456

inertia of the vehicle body has little or effect on the vertical displacement of the

vehicle body under free-vibration. Next, the author presents the rolling and pitching

motion of the three different vehicle bodies, respectively, in Figure G.14. Again, the

reader can see that the vehicle body simulated by beam elements, which can have

different mass moments of inertia, has little or no effect in the finite element solution.

Therefore, the equivalent mass moment of inertia has little or no dynamic effect when

one models the vehicle body and bogies using beam elements in the finite element

solution.

-0.20

-0.15

-0.10

-0.05

0.00

0.0 0.5 1.0 1.5 2.0 2.5 3.0

Time (sec)

Ve

rtic

al D

isp

lac

em

en

t (m

)

Model A Model B Model C

Figure G.13: Vertical displacement of the vehicle body as a function of time

-0.08

-0.06

-0.04

-0.02

0.00

0.02

0.04

0.06

0.08

0.0 0.5 1.0 1.5 2.0 2.5 3.0

Time (sec)

Ro

llin

g D

isp

lac

em

en

t (r

ad

)


(a)


457

-0.010

-0.008

-0.005

-0.003

0.000

0.003

0.005

0.008

0.010

0.0 0.5 1.0 1.5 2.0 2.5 3.0

Time (sec)

Pit

ch

ing

Dis

pla

ce

me

nt

(ra

d)


Figure G.14: (a) Rolling; (b) pitching motion of vehicle body as a function of time

using the three different mass moments of inertia for the vehicle body

It should be noted that the centre of gravity of the bogie and wheels as well as its mass

moment of inertia are computed in a similar manner.

G.6 Modal analysis of the railway vehicles

Table G.2 presents results obtained from a modal analysis of the 201 Class

locomotive and Mark3 railway coach. The different mode shapes of the railway

vehicles are then shown in Figure G.15.

Table G.2: Modal analysis of vehicle body of the 201 Class locomotive and Mark3

railway coach used extensively throughout this thesis

Mode Frequency (Hz)

201 Loco (V1) MK3 Coach (V2)

Rolling 1.917 2.003

Bouncing 5.433 6.255

Pitching 5.968 7.129

Yawing 54.469 51.761

(b)


458

Figure G.15: Six modes that the railway vehicle can experience as it travels along

the rail (a) longitudinal oscillation; (b) bouncing; (c) lateral

oscillation; (d) rolling; (e) yawing; (f) pitching (Iwnicki, 2006)

G.7 Braking and accelerating effects of a train

In Section 4.4.4, the train experiences acceleration; thus, the speed of the train is no

longer constant. The mathematical formula used to compute the change in speed and

the positions of any wheel at any time t are derived in this sub-section. One begins

with a linear acceleration defined by:

( )( )

2

2

d X tX t Bt D

dt= = +&& (G.3)

y

x

y

z

Longitudinal oscillation Rolling

x

y

x

z

Yawing Bouncing

x

z

Lateral oscillation

x

y

Pitching

(c)

(a)

(b) (e)

(f)

(d)


459

where ( )X t and ( )X t&& are the displacement and acceleration of the front wheel of the

train at time t, and B and D are unknown constants. At time 0,t = on finds that

Equation (G.3) reduces to:

( ) ( )0 0X B D= +&&

( )0X D=&& (G.4)

Substituting Equation (G.4) back into (G.3) and solving for B gives:

( ) ( )0X t Bt X= +&& && (G.5a)

( ) ( )0X t XB

t

−=

&& &&

(G.5b)

Integrating Equation (G.3), one finds the velocity of wheel at time t equal to:

( )( )

0 0

2

2

t t

t t

d X tdt Bt D dt

dt= +∫ ∫ (G.6a)

( )

00

2 2

2 2

t t

tt

d X t BtDt

dt

= +

(G.6b)

( ) ( ) 220 0

02 2

dX t dX t BtBtDt Dt

dt dt

− = + − −

(G.6c)

( ) ( ) 220 0

02 2

dX t dX t BtBtDt Dt

dt dt= + + − − (G.6d)

where 0t is the initial time.


460

Integrating Equation (G.6d), one finds the displacement of wheel at time t equal to:

( ) ( )

0 0

220 0

02 2

t t

t t

dX t dX t BtBtdt Dt Dt dt

dt dt

= + + − −

∫ ∫ (G.7a)

( )( )

0

0

23 20 0

06 2 2

t

t

t

t

dX t Bt tBt DtX t t Dt t

dt

= + + − −

(G.7b)

( ) ( )( ) 23 2

0 00 0

6 2 2

dX t Bt tBt DtX t X t t Dt t

dt

− = + + − −

( ) 3 2 30 20 0 0

0 06 2 2

dX t Bt Dt Btt Dt

dt

− − − + +

(G.7c)

( ) ( )( ) 23 2

0 00 0

6 2 2

dX t Bt tBt DtX t X t t Dt t

dt= + + + − −

( ) 3 2 30 20 0 0

0 06 2 2

dX t Bt Dt Btt Dt

dt− − − + + (G.7d)

If one assumes that time 0 0t = and that the acceleration remains constant, 0B = , then

Equations (G.6d) can be rewritten as follows:

( ) ( ) ( )( )

0 00

dX t dX dXDt X t

dt dt dt= + = + && (G.8a)

or

( ) ( ) ( )0 0X t X X t= +& & && (G.8b)

Equally, if 0 0t = and 0B = , Equation (G.7d) can be rewritten as follows:

( ) ( )( )

( )( ) ( ) 220 0 0

0 02 2

dX dX X tDtX t X t X t

dt dt= + + = + +

&&

(G.9a)

or

( ) ( ) ( ) ( ) 210 0 0

2X t X X t X t= + +& && (G.9b)


461

With a change of notation in Equation (G.8b) and (G.9b), one can write the velocity

and displacement of the front wheel of the train as follows:

( )X t c at= +& (G.10a)

( ) ( ) 210

2X t X ct at= + + (G.10b)

If one assumes that the front wheel of the train is at the left support of the bridge at

time 0t = , then Equation (G.10b) can be rewritten as follows:

( ) 21

2X t ct at= + (G.10c)

Using Equation (G.10c) and (3.39), the position of the m-th wheel, of the k-th bogie,

and the j-th carriage of the train is then equal to:

( ) ( ) ( ) ( )2

, ,

11 1 1

2w w wm j k

X t ct at D j B k A m= + − − − − − − (G.11)

As an example, a train is assumed to have an initial speed c of 45.55 m/s (164 km/hr)

at time t = 0 sec and the moment the front wheel enters the Boyne Viaduct Railway

Bridge, it begins to decelerate at a = -2.075 m/s2, so that the train comes to a complete

stop after 500 m. Using Equation (G.10c), one is first able to compute the time when

the front wheel arrives at a certain position on the bridge, while Equation (G.10a) is

then used to compute the train speed at that particular time. In Table G.3, the author

selects different positions along the bridge and computes the time and velocity of the

front wheel at these locations. Assuming the train consists of a single 201 Class


462

locomotive and three Mark3 railway coaches, then according to Section G.2, the

distance from the front wheel to the rear wheel of the train is 85.882 m; thus, the front

wheel of the train would have to travel 80.77 85.88 166.65 m+ = before the rear wheel

of the train reaches the right hand support of the bridge.

Table G.3: Wheel position and velocity of a train as it decelerates along Boyne Bridge



(s) (m) (m/s)





As a second example, the train is now assumed to be stationary with the front wheel

of the train at the left-hand support at time t = 0 sec. The train then accelerates

constantly at a = 2.075 m/s2, so that it reaches a speed of 45.55 m/s (164 km/hr) after

500 m. Using the same formulae as before, the author computes the time it takes for

the front wheel of the train to reach a specific location on the bridge as well as the

velocity of the train at that particular time as presented in Table G.4

Table G.4: Wheel position and velocity of a train as it accelerates along Boyne Bridge



(s) (m) (m/s)





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